Loading [MathJax]/jax/element/mml/optable/BasicLatin.js
Research article

Ries impact deposits in the North Alpine Foreland Basin of Germany as a terrestrial analog site for impact-produced seismites and sand spikes on planet Mars

  • The North Alpine foreland basin (NAFB) in Germany is characterized by various types of sedimentologic features that make it an excellent terrestrial analog of regions affected by high-energy asteroid impact-quakes on Mars. Impact events have shaped all planetary bodies in the inner Solar System over the past >4 Gyr. The well-preserved Ries impact crater (Baden-Württemberg and Bavaria), formed around 14.8 Ma, has recently been linked to an earthquake-produced seismite horizon in Mid-Miocene NAFB sediments that exhibits typical dewatering structures and is associated with sand spikes, seismically produced pin-shaped pseudo-concretions. In this terrestrial setting, the sand spike tails systematically point away from the Ries crater. On its path across Gale Crater, the Mars rover Curiosity seems to have observed a similar seismite horizon in early Hesperian lacustrine deposits including clastic dikes, convolute bedding, and, likely, sand spikes. Their orientation suggests that the nearby Slagnos impact crater might be the seismic source for the formation of those seismites. The Ries impact–seismite deposits can be traced over a distance of more than 200 km from the source crater (northern Switzerland), which makes the NAFB an excellent terrestrial analog for similar deposits and their sedimentologic inventory within Gale Crater's lake deposits on Mars.

    Citation: Elmar Buchner, Volker J Sach, Martin Schmieder. Ries impact deposits in the North Alpine Foreland Basin of Germany as a terrestrial analog site for impact-produced seismites and sand spikes on planet Mars[J]. AIMS Geosciences, 2025, 11(1): 68-90. doi: 10.3934/geosci.2025005

    Related Papers:

    [1] Zhi-Wei Sun . New series for powers of π and related congruences. Electronic Research Archive, 2020, 28(3): 1273-1342. doi: 10.3934/era.2020070
    [2] Harman Kaur, Meenakshi Rana . Congruences for sixth order mock theta functions λ(q) and ρ(q). Electronic Research Archive, 2021, 29(6): 4257-4268. doi: 10.3934/era.2021084
    [3] Jorge Garcia Villeda . A computable formula for the class number of the imaginary quadratic field Q(p), p=4n1. Electronic Research Archive, 2021, 29(6): 3853-3865. doi: 10.3934/era.2021065
    [4] Fedor Petrov, Zhi-Wei Sun . Proof of some conjectures involving quadratic residues. Electronic Research Archive, 2020, 28(2): 589-597. doi: 10.3934/era.2020031
    [5] Jin-Yun Guo, Cong Xiao, Xiaojian Lu . On n-slice algebras and related algebras. Electronic Research Archive, 2021, 29(4): 2687-2718. doi: 10.3934/era.2021009
    [6] Dušan D. Repovš, Mikhail V. Zaicev . On existence of PI-exponents of unital algebras. Electronic Research Archive, 2020, 28(2): 853-859. doi: 10.3934/era.2020044
    [7] Victor J. W. Guo . A family of q-congruences modulo the square of a cyclotomic polynomial. Electronic Research Archive, 2020, 28(2): 1031-1036. doi: 10.3934/era.2020055
    [8] Chen Wang . Two congruences concerning Apéry numbers conjectured by Z.-W. Sun. Electronic Research Archive, 2020, 28(2): 1063-1075. doi: 10.3934/era.2020058
    [9] Dmitry Krachun, Zhi-Wei Sun . On sums of four pentagonal numbers with coefficients. Electronic Research Archive, 2020, 28(1): 559-566. doi: 10.3934/era.2020029
    [10] Hai-Liang Wu, Zhi-Wei Sun . Some universal quadratic sums over the integers. Electronic Research Archive, 2019, 27(0): 69-87. doi: 10.3934/era.2019010
  • The North Alpine foreland basin (NAFB) in Germany is characterized by various types of sedimentologic features that make it an excellent terrestrial analog of regions affected by high-energy asteroid impact-quakes on Mars. Impact events have shaped all planetary bodies in the inner Solar System over the past >4 Gyr. The well-preserved Ries impact crater (Baden-Württemberg and Bavaria), formed around 14.8 Ma, has recently been linked to an earthquake-produced seismite horizon in Mid-Miocene NAFB sediments that exhibits typical dewatering structures and is associated with sand spikes, seismically produced pin-shaped pseudo-concretions. In this terrestrial setting, the sand spike tails systematically point away from the Ries crater. On its path across Gale Crater, the Mars rover Curiosity seems to have observed a similar seismite horizon in early Hesperian lacustrine deposits including clastic dikes, convolute bedding, and, likely, sand spikes. Their orientation suggests that the nearby Slagnos impact crater might be the seismic source for the formation of those seismites. The Ries impact–seismite deposits can be traced over a distance of more than 200 km from the source crater (northern Switzerland), which makes the NAFB an excellent terrestrial analog for similar deposits and their sedimentologic inventory within Gale Crater's lake deposits on Mars.



    The classical rational Ramanujan-type series for π1 (cf. [1,2,8,27] and a nice introduction by S. Cooper [10,Chapter 14]) have the form

    k=0bk+cmka(k)=λdπ,()

    where b,c,m are integers with bm0, d is a positive squarefree number, λ is a nonzero rational number, and a(k) is one of the products

    (2kk)3, (2kk)2(3kk), (2kk)2(4k2k), (2kk)(3kk)(6k3k).

    In 1997 Van Hamme [47] conjectured that such a series () has a p-adic analogue of the form

    p1k=0bk+cmka(k)cp(εddp) (mod p3),

    where p is any odd prime with pdm and λZp, ε1{±1} and εd=1 if d>1. (As usual, Zp denotes the ring of all p-adic integers, and (p) stands for the Legendre symbol.) W. Zudilin [53] followed Van Hamme's idea to provide more concrete examples. Sun [33] realized that many Ramanujan-type congruences are related to Bernoulli numbers or Euler numbers. In 2016 the author [44] thought that all classical Ramanujan-type congruences have their extensions like

    pn1k=0(21k+8)(2kk)3pn1k=0(21k+8)(2kk)3(pn)3(2nn)3Zp,

    where p is an odd prime, and nZ+={1,2,3,}. See Sun [45,Conjectures 21-24] for more such examples and further refinements involving Bernoulli or Euler numbers.

    During the period 2002–2010, some new Ramanujan-type series of the form () with a(k) not a product of three nontrivial parts were found (cf. [3,4,9,29]). For example, H. H. Chan, S. H. Chan and Z. Liu [3] proved that

    n=05n+164nDn=83π,

    where Dn denotes the Domb number nk=0(nk)2(2kk)(2(nk)nk); Zudilin [53] conjectured its p-adic analogue:

    p1k=05k+164kDkp(p3) (mod p3)for any prime p>3.

    The author [45,Conjecture 77] conjectured further that

    1(pn)3(pn1k=05k+164kDk(p3)pn1k=05k+164rDk)Zp

    for each odd prime p and positive integer n.

    Let b,cZ. For each nN={0,1,2,}, we denote the coefficient of xn in the expansion of (x2+bx+c)n by Tn(b,c), and call it a generalized central trinomial coefficient. In view of the multinomial theorm, we have

    Tn(b,c)=n/2k=0(n2k)(2kk)bn2kck=n/2k=0(nk)(nkk)bn2kck.

    Note also that

    T0(b,c)=1,  T1(b,c)=b,

    and

    (n+1)Tn+1(b,c)=(2n+1)bTn(b,c)n(b24c)Tn1(b,c)

    for all nZ+. Clearly, Tn(2,1)=(2nn) for all nN. Those Tn:=Tn(1,1) with nN are the usual central trinomial coefficients, and they play important roles in enumerative combinatorics. We view Tn(b,c) as a natural generalization of central binomial and central trinomial coefficients.

    For nN the Legendre polynomial of degree n is defined by

    Pn(x):=nk=0(nk)(n+kk)(x12)k.

    It is well-known that if b,cZ and b24c0 then

    Tn(b,c)=(b24c)nPn(bb24c)for all nN.

    Via the Laplace-Heine asymptotic formula for Legendre polynomials, for any positive real numbers b and c we have

    Tn(b,c)(b+2c)n+1/224cnπas n+

    (cf. [40]). For any real numbers b and c<0, S. Wagner [48] confirmed the author's conjecture that

    lim

    In 2011, the author posed over 60 conjectural series for 1/\pi of the following new types with a,b,c,d,m integers and mbcd(b^2-4c) nonzero (cf. Sun [34,40]).

    Type Ⅰ. \sum_{k = 0}^\infty \frac{a+dk}{m^k} \binom{2k}k^2T_k(b,c) .

    Type Ⅱ. \sum_{k = 0}^\infty \frac{a+dk}{m^k} \binom{2k}k \binom{3k}kT_k(b,c) .

    Type Ⅲ. \sum_{k = 0}^\infty \frac{a+dk}{m^k} \binom{4k}{2k} \binom{2k}kT_k(b,c) .

    Type Ⅳ. \sum_{k = 0}^\infty \frac{a+dk}{m^k} \binom{2k}{k}^2T_{2k}(b,c) .

    Type Ⅴ. \sum_{k = 0}^\infty \frac{a+dk}{m^k} \binom{2k}{k} \binom{3k}kT_{3k}(b,c) .

    Type Ⅵ. \sum_{k = 0}^\infty \frac{a+dk}{m^k}T_{k}(b,c)^3,

    Type Ⅶ. \sum_{k = 0}^\infty \frac{a+dk}{m^k} \binom{2k}kT_{k}(b,c)^2,

    In general, the corresponding p -adic congruences of these seven-type series involve linear combinations of two Legendre symbols. The author's conjectural series of types Ⅰ-Ⅴ and Ⅶ were studied in [6,49,54]. The author's three conjectural series of type Ⅵ and two series of type Ⅶ remain open. For example, the author conjectured that

    \sum\limits_{k = 0}^\infty \frac{3990k+1147}{(-288)^{3k}}T_k(62,95^2)^3 = \frac{432}{95\pi}(94\sqrt2+195\sqrt{14})

    as well as its p -adic analogue

    \sum\limits_{k = 0}^{p-1} \frac{3990k+1147}{(-288)^{3k}}T_k(62,95^2)^3{\equiv} \frac p{19} \left(4230 \left( \frac{-2}p \right)+17563 \left( \frac{-14}p \right) \right)\ ({\rm{mod}}\ {p^2}),

    where p is any prime greater than 3 .

    In 1905, J. W. L. Glaisher [15] proved that

    \sum\limits_{k = 0}^\infty \frac{(4k-1) \binom{2k}k^4}{(2k-1)^4256^k} = - \frac 8{\pi^2}.

    This actually follows from the following finite identity observed by the author [38]:

    \sum\limits_{k = 0}^n \frac{(4k-1) \binom{2k}k^4}{(2k-1)^4256^k} = -(8n^2+4n+1) \frac{ \binom{2n}n^4}{256^n}\quad\ \text{for all}\ n\in{\Bbb N}.

    Motivated by Glaisher's identity and Ramanujan-type series for 1/\pi , we obtain the following theorem.

    Theorem 1.1. We have the following identities:

    \begin{align} \sum\limits_{k = 0}^\infty \frac{k(4k-1) \binom{2k}k^3}{(2k-1)^2(-64)^k}& = - \frac1{\pi}, \end{align} (1.1)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(4k-1) \binom{2k}k^3}{(2k-1)^3(-64)^k}& = \frac2{\pi}, \end{align} (1.2)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(12k^2-1) \binom{2k}k^3}{(2k-1)^2 256^k}& = - \frac2{\pi}, \end{align} (1.3)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{k(6k-1) \binom{2k}k^3}{(2k-1)^3256^k}& = \frac1{2\pi}, \end{align} (1.4)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(28k^2-4k-1) \binom{2k}k^3}{(2k-1)^2(-512)^k}& = - \frac{3\sqrt2}{\pi}, \end{align} (1.5)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(30k^2+3k-2) \binom{2k}k^3}{(2k-1)^3(-512)^k}& = \frac{27\sqrt2}{8\pi}, \end{align} (1.6)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(28k^2-4k-1) \binom{2k}k^3}{(2k-1)^2 4096^k}& = - \frac3{\pi}, \end{align} (1.7)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(42k^2-3k-1) \binom{2k}k^3}{(2k-1)^3 4096^k}& = \frac{27}{8\pi}, \end{align} (1.8)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(34k^2-3k-1) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)(-192)^k}& = - \frac{10}{\sqrt3\,\pi}, \end{align} (1.9)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(64k^2-11k-7) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)(-192)^k}& = - \frac{125\sqrt3}{9\pi}, \end{align} (1.10)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(14k^2+k-1) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)216^k}& = - \frac{\sqrt3}{\pi}, \end{align} (1.11)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(90k^2+7k+1) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)216^k}& = \frac{9\sqrt3}{2\pi}, \end{align} (1.12)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(34k^2-3k-1) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)(-12)^{3k}}& = - \frac{2\sqrt3}{\pi}, \end{align} (1.13)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(17k+5) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)(-12)^{3k}}& = \frac{9\sqrt3}{\pi}, \end{align} (1.14)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(111k^2-7k-4) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)1458^k}& = - \frac{45}{4\pi}, \end{align} (1.15)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(1524k^2+899k+263) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)1458^k}& = \frac{3375}{4\pi}, \end{align} (1.16)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(522k^2-55k-13) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)(-8640)^k}& = - \frac{54\sqrt{15}}{5\pi}, \end{align} (1.17)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(1836k^2+2725k+541) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)(-8640)^k}& = \frac{2187\sqrt{15}}{5\pi}, \end{align} (1.18)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(529k^2-45k-16) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)15^{3k}}& = - \frac{55\sqrt3}{2\pi}, \end{align} (1.19)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(77571k^2+68545k+16366) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)15^{3k}}& = \frac{59895\sqrt3}{2\pi}, \end{align} (1.20)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(574k^2-73k-11) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)(-48)^{3k}}& = -20 \frac{\sqrt3}{\pi}, \end{align} (1.21)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(8118k^2+9443k+1241) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)(-48)^{3k}}& = \frac{2250\sqrt3}{\pi}, \end{align} (1.22)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(978k^2-131k-17) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)(-326592)^k}& = - \frac{990\sqrt7}{49\pi}, \end{align} (1.23)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(592212k^2+671387k^2+77219) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)(-326592)^k}& = \frac{4492125\sqrt7}{49\pi}, \end{align} (1.24)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(116234k^2-17695k-1461) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)(-300)^{3k}}& = -2650 \frac{\sqrt3}{\pi}, \end{align} (1.25)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(223664832k^2+242140765k+18468097) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)(-300)^{3k}}& = 33497325 \frac{\sqrt3}{\pi}, \end{align} (1.26)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(122k^2+3k-5) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)648^k}& = - \frac{21}{2\pi}, \end{align} (1.27)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(1903k^2+114k+41) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)648^k}& = \frac{343}{2\pi}, \end{align} (1.28)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(40k^2-2k-1) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)(-1024)^k}& = - \frac4{\pi}, \end{align} (1.29)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(8k^2-2k-1) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)(-1024)^k}& = - \frac{16}{5\pi}, \end{align} (1.30)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(176k^2-6k-5) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)48^{2k}}& = -8 \frac{\sqrt3}{\pi}, \end{align} (1.31)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(208k^2+66k+23) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)48^{2k}}& = \frac{128}{\sqrt3\,\pi}, \end{align} (1.32)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(6722k^2-411k-152) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)(-63^2)^k}& = -195 \frac{\sqrt7}{\pi}, \end{align} (1.33)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(281591k^2-757041k-231992) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)(-63^2)^k}& = -274625 \frac{\sqrt7}{\pi}, \end{align} (1.34)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(560k^2-42k-11) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)12^{4k}}& = -24 \frac{\sqrt2}{\pi}, \end{align} (1.35)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(112k^2+114k+23) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)12^{4k}}& = \frac{256\sqrt2}{5\pi}, \end{align} (1.36)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(248k^2-18k-5) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)(-3\times2^{12})^k}& = - \frac{28}{\sqrt3\,\pi}, \end{align} (1.37)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(680k^2+1482k+337) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)(-3\times2^{12})^k}& = \frac{5488\sqrt3}{9\pi}, \end{align} (1.38)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(1144k^2-102k-19) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)(-2^{10}3^4)^k}& = - \frac{60}{\pi}, \end{align} (1.39)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(3224k^2+4026k+637) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)(-2^{10}3^4)^k}& = \frac{2000}{\pi}, \end{align} (1.40)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(7408k^2-754k-103) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)28^{4k}}& = - \frac{560\sqrt3}{3\pi}, \end{align} (1.41)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(3641424k^2+4114526k+493937) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)28^{4k}}& = 896000 \frac{\sqrt3}{\pi}, \end{align} (1.42)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(4744k^2-534k-55) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)(-2^{14}3^45)^k}& = - \frac{1932\sqrt5}{25\pi}, \end{align} (1.43)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(18446264k^2+20356230k+1901071) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)(-2^{14}3^45)^k}& = 66772496 \frac{\sqrt5}{25\pi}, \end{align} (1.44)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(413512k^2-50826k-3877) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)(-2^{10}21^4)^k}& = - \frac{12180}{\pi}, \end{align} (1.45)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(1424799848k^2+1533506502k+108685699) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)(-2^{10}21^4)^k}& = \frac{341446000}{\pi}, \end{align} (1.46)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(71312k^2-7746k-887) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)1584^{2k}}& = -840 \frac{\sqrt{11}}{\pi}, \end{align} (1.47)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(50678512k^2+56405238k+5793581) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)1584^{2k}}& = 5488000 \frac{\sqrt{11}}{\pi}, \end{align} (1.48)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(7329808k^2-969294k-54073) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)396^{4k}}& = -120120 \frac{\sqrt2}{\pi}, \end{align} (1.49)
    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^\infty \frac{(2140459883152k^2+2259867244398k+119407598201) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)396^{4k}} \\&\qquad\qquad\qquad = 44\times1820^3 \frac{\sqrt2}{\pi}, \end{aligned} \end{equation} (1.50)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(164k^2-k-3) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)20^{3k}}& = - \frac{7\sqrt5}{2\pi}, \end{align} (1.51)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(2696k^2+206k+93) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)20^{3k}} & = \frac{686}{\sqrt5\,\pi}, \end{align} (1.52)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(220k^2-8k-3) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(-2^{15})^k} & = - \frac{7\sqrt2}{\pi}, \end{align} (1.53)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(836k^2-1048k-309) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(-2^{15})^k} & = - \frac{686\sqrt2}{\pi}, \end{align} (1.54)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(504k^2-11k-8) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(-15)^{3k}} & = - \frac{9\sqrt{15}}{\pi}, \end{align} (1.55)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(189k^2-11k-8) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(-15)^{3k}}& = - \frac{243\sqrt{15}}{35\pi}, \end{align} (1.56)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(516k^2-19k-7) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(2\times30^3)^k} & = - \frac{11\sqrt{15}}{2\pi}, \end{align} (1.57)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(3237k^2+1922k+491) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(2\times30^3)^k} & = \frac{3993\sqrt{15}}{10\pi}, \end{align} (1.58)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(684k^2-40k-7) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(-96)^{3k}}& = - \frac{9\sqrt6}{\pi}, \end{align} (1.59)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(2052k^2+2536k+379) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(-96)^{3k}}& = \frac{486\sqrt6}{\pi}, \end{align} (1.60)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(2556k^2-131k-29) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)66^{3k}}& = - \frac{63\sqrt{33}}{4\pi}, \end{align} (1.61)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(203985k^2+212248k+38083) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)66^{3k}}& = \frac{83349\sqrt{33}}{4\pi}, \end{align} (1.62)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(5812k^2-408k-49) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(-3\times160^3)^k}& = - \frac{253\sqrt{30}}{9\pi}, \end{align} (1.63)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(3471628k^2+3900088k+418289) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(-3\times160^3)^k}& = \frac{32388554\sqrt{30}}{135\pi}, \end{align} (1.64)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(35604k^2-2936k-233) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(-960)^{3k}}& = -189 \frac{\sqrt{15}}{\pi}, \end{align} (1.65)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(13983084k^2+15093304k+1109737) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(-960)^{3k}}& = \frac{4500846\sqrt{15}}{5\pi}, \end{align} (1.66)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(157752k^2-11243k-1304) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)255^{3k}}& = - \frac{513\sqrt{255}}{2\pi}, \end{align} (1.67)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(28240947k^2+31448587k+3267736) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)255^{3k}}& = \frac{45001899\sqrt{255}}{70\pi}, \end{align} (1.68)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{(2187684k^2-200056k-11293) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(-5280)^{3k}} & = -1953 \frac{\sqrt{330}}{\pi}, \end{align} (1.69)
    \begin{equation} \begin{aligned} &\sum\limits_{k = 0}^\infty \frac{(101740699836k^2+107483900696k+5743181813) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(-5280)^{3k}} \\&\qquad\qquad\qquad = \frac{4966100118\sqrt{330}}{5\pi}, \end{aligned} \end{equation} (1.70)
    \begin{equation} \begin{aligned} &\sum\limits_{k = 0}^\infty \frac{(16444841148k^2-1709536232k-53241371) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(-640320)^{3k}} \\&\qquad\qquad\qquad = -1672209 \frac{\sqrt{10005}}{\pi}, \end{aligned} \end{equation} (1.71)

    and

    \begin{equation} \begin{aligned} \sum\limits_{k = 0}^\infty \frac{P(k) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(-640320)^{3k}} = \frac{18\times557403^3\sqrt{10005}}{5\pi}, \end{aligned} \end{equation} (1.72)

    where

    \begin{align*} P(k): = &637379600041024803108 k^2 + 657229991696087780968 k \\&+ 19850391655004126179. \end{align*}

    Recall that the Catalan numbers are given by

    C_n: = \frac{ \binom{2n}n}{n+1} = \binom{2n}n- \binom{2n}{n+1}\ \ (n\in{\Bbb N}).

    For k\in{\Bbb N} it is easy to see that

    \frac{ \binom{2k}k}{2k-1} = \begin{cases}-1& \text{if}\ k = 0,\\2C_{k-1}& \text{if}\ k > 0.\end{cases}

    Thus, for any a,b,c,m\in{\Bbb Z} with |m|{\geq}64 , we have

    \begin{align*} \sum\limits_{k = 0}^\infty \frac{(ak^2+bk+c) \binom{2k}k^3}{(2k-1)^3m^k} = &-c+\sum\limits_{k = 1}^\infty \frac{(ak^2+bk+c)(2C_{k-1})^3}{m^k} \\ = &-c+ \frac 8m\sum\limits_{k = 0}^\infty \frac{a(k+1)^2+b(k+1)+c}{m^k}C_k^3. \end{align*}

    For example, (1.2) has the equivalent form

    \sum\limits_{k = 0}^\infty \frac{4k+3}{(-64)^k}C_k^3 = 8- \frac{16}{\pi}.\;\;\;\;\;\;\;\;(1.2')

    For any odd prime p , the congruence (1.4) of V.J.W. Guo and J.-C. Liu [19] has the equivalent form

    \sum\limits_{k = 0}^{(p+1)/2} \frac{(4k-1) \binom{2k}k^3}{(2k-1)^3(-64)^k}{\equiv} p \left( \frac{-1}p \right)+p^3(E_{p-3}-2)\ ({\rm{mod}}\ {p^4})

    (where E_0,E_1,\ldots are the Euler numbers), and we note that this is also equivalent to the congruence

    \sum\limits_{k = 0}^{(p-1)/2} \frac{4k+3}{(-64)^k}C_k^3{\equiv}8 \left(1-p \left( \frac{-1}p \right)-p^3(E_{p-3}-2) \right)\ ({\rm{mod}}\ {p^4}).

    Recently, C. Wang [50] proved that for any prime p>3 we have

    \sum\limits_{k = 0}^{(p+1)/2} \frac{(3k-1) \binom{2k}k^3}{(2k-1)^216^k}{\equiv} p+2p^3 \left( \frac{-1}p \right)(E_{p-3}-3)\ ({\rm{mod}}\ {p^4})

    and

    \sum\limits_{k = 0}^{p-1} \frac{(3k-1) \binom{2k}k^3}{(2k-1)^216^k}{\equiv} p-2p^3\ ({\rm{mod}}\ {p^4}).

    (Actually, Wang stated his results only in the language of hypergeometric series.) These two congruences extend a conjecture of Guo and M. J. Schlosser [21].

    We are also able to prove some other variants of Ramanujan-type series such as

    \sum\limits_{k = 0}^\infty \frac{(56k^2+118k+61) \binom{2k}k^3}{(k+1)^24096^k} = \frac{192}{\pi}

    and

    \sum\limits_{k = 0}^\infty \frac{(420k^2+992k+551) \binom{2k}k^3}{(k+1)^2(2k-1)4096^k} = - \frac{1728}{\pi}.

    Now we state our second theorem.

    Theorem 1.2. We have the identities

    \begin{align} \sum\limits_{k = 1}^\infty \frac{28k^2+31k+8}{(2k+1)^2k^3 \binom{2k}k^3}& = \frac{\pi^2-8}2, \end{align} (1.73)
    \begin{align} \sum\limits_{k = 1}^\infty \frac{42k^2+39k+8}{(2k+1)^3k^3 \binom{2k}k^3}& = \frac{9\pi^2-88}2, \end{align} (1.74)
    \begin{align} \sum\limits_{k = 1}^\infty \frac{(8k^2+5k+1)(-8)^k}{(2k+1)^2k^3 \binom{2k}k^3}& = 4-6G, \end{align} (1.75)
    \begin{align} \sum\limits_{k = 1}^\infty \frac{(30k^2+33k+7)(-8)^k}{(2k+1)^3k^3 \binom{2k}k^3}& = 54G-52, \end{align} (1.76)
    \begin{align} \sum\limits_{k = 1}^\infty \frac{(3k+1)16^k}{(2k+1)^2k^3 \binom{2k}k^3}& = \frac{\pi^2-8}2, \end{align} (1.77)
    \begin{align} \sum\limits_{k = 1}^\infty \frac{(4k+1)(-64)^k}{(2k+1)^2k^2 \binom{2k}k^3}& = 4-8G, \end{align} (1.78)
    \begin{align} \sum\limits_{k = 1}^\infty \frac{(4k+1)(-64)^k}{(2k+1)^3k^3 \binom{2k}k^3}& = 16G-16, \end{align} (1.79)
    \begin{align} \sum\limits_{k = 1}^\infty \frac{(2k^2-11k-3)8^k}{(2k+1)(3k+1)k^3 \binom{2k}k^2 \binom{3k}k}& = \frac{48-5\pi^2}2, \end{align} (1.80)
    \begin{align} \sum\limits_{k = 2}^\infty \frac{(178k^2-103k-39)8^k}{(k-1)(2k+1)(3k+1)k^3 \binom{2k}k^2 \binom{3k}k}& = \frac{1125\pi^2-11096}{36}, \end{align} (1.81)
    \begin{align} \sum\limits_{k = 1}^\infty \frac{(5k+1)(-27)^k}{(2k+1)(3k+1)k^2 \binom{2k}k^2 \binom{3k}k}& = 6-9K, \end{align} (1.82)
    \begin{align} \sum\limits_{k = 2}^\infty \frac{(45k^2+5k-2)(-27)^{k-1}}{(k-1)(2k+1)(3k+1)k^3 \binom{2k}k^2 \binom{3k}k}& = \frac{37-48K}{16}, \end{align} (1.83)
    \begin{align} \sum\limits_{k = 1}^\infty \frac{(98k^2-21k-8)81^k}{(2k+1)(4k+1)k^3 \binom{2k}k^2 \binom{4k}{2k}}& = 216-20\pi^2, \end{align} (1.84)
    \begin{align} \sum\limits_{k = 2}^\infty \frac{(1967k^2-183k-104)81^k}{(k-1)(2k+1)(4k+1)k^3 \binom{2k}k^2 \binom{4k}{2k}}& = \frac{20000\pi^2-190269}{120}, \end{align} (1.85)
    \begin{align} \sum\limits_{k = 1}^\infty \frac{(46k^2+3k-1)(-144)^k}{(2k+1)(4k+1)k^3 \binom{2k}k^2 \binom{4k}{2k}}& = 72- \frac{225}2K, \end{align} (1.86)
    \begin{align} \sum\limits_{k = 2}^\infty \frac{(343k^2+18k-16)(-144)^k}{(k-1)(2k+1)(4k+1)k^3 \binom{2k}k^2 \binom{4k}{2k}}& = \frac{9375K-7048}{10}, \end{align} (1.87)

    where

    G: = \sum\limits_{k = 0}^\infty \frac{(-1)^k}{(2k+1)^2}\ \ \mathit{\text{and}}\ \ K: = \sum\limits_{k = 0}^\infty \frac{( \frac k3)}{k^2}.

    For k = j+1\in{\Bbb Z}^+ , it is easy to see that

    (k-1)k \binom{2k}k = 2(2j+1)j \binom{2j}j.

    Thus, for any a,b,c,m\in{\Bbb Z} with 0<|m| \leq 64 , we have

    \sum\limits_{j = 1}^\infty \frac{(aj^2+bj+c)m^j}{(2j+1)^3j^3 \binom{2j}j^3} = \frac 8m\sum\limits_{k = 2}^\infty \frac{(a(k-1)^2+b(k-1)+c)m^k}{(k-1)^3k^3 \binom{2k}k^3}.

    For example, (1.77) has the following equivalent form

    \sum\limits_{k = 2}^\infty \frac{(2k-1)(3k-2)16^k}{(k-1)^3k^3 \binom{2k}k^3} = \pi^2-8.\;\;\;\;\;\;\;\;\;\;\;(1.77')

    In contrast with the Domb numbers, we introduce a new kind of numbers

    S_n: = \sum\limits_{k = 0}^n \binom nk^2T_kT_{n-k}\ \ (n = 0,1,2,\ldots).

    The values of S_n\ (n = 0,\ldots,10) are

    1,\, 2,\, 10,\, 68,\, 586,\, 5252,\, 49204,\, 475400,\, 4723786,\, 47937812,\, 494786260

    respectively. We may extend the numbers S_n\ (n\in{\Bbb N}) further. For b,c\in{\Bbb Z} , we define

    S_n(b,c): = \sum\limits_{k = 0}^n \binom nk^2T_k(b,c)T_{n-k}(b,c)\ \ (n = 0,1,2,\ldots).

    Note that S_n(1,1) = S_n and S_n(2,1) = D_n for all n\in{\Bbb N} .

    Now we state our third theorem.

    Theorem 1.3. We have

    \begin{align} \sum\limits_{k = 0}^\infty \frac{7k+3}{24^k}S_k(1,-6)& = \frac{15}{\sqrt2\,\pi}, \end{align} (1.88)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{12k+5}{(-28)^k}S_k(1,7)& = \frac{6\sqrt7}{\pi}, \end{align} (1.89)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{84k+29}{80^k}S_k(1,-20)& = \frac{24\sqrt{15}}{\pi}, \end{align} (1.90)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{3k+1}{(-100)^k}S_k(1,25)& = \frac{25}{8\pi}, \end{align} (1.91)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{228k+67}{224^k}S_k(1,-56)& = \frac{80\sqrt7}{\pi}, \end{align} (1.92)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{399k+101}{(-676)^k}S_k(1,169)& = \frac{2535}{8\pi}, \end{align} (1.93)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{2604k+563}{2600^k}S_k(1,-650)& = \frac{850\sqrt{39}}{3\pi}, \end{align} (1.94)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{39468k+7817}{(-6076)^k}S_k(1,1519)& = \frac{4410\sqrt{31}}{\pi}, \end{align} (1.95)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{41667k+7879}{9800^k}S_k(1,-2450)& = \frac{40425\sqrt6}{4\pi}, \end{align} (1.96)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{74613k+10711}{(-530^2)^k}S_k(1,265^2)& = \frac{1615175}{48\pi}. \end{align} (1.97)

    Remark 1.1. The author found the 10 series in Theorem 1.3 in Nov. 2019.

    We shall prove Theorems 1.1-1.3 in the next section. In Sections 3-10, we propose 117 new conjectural series for powers of \pi involving generalized central trinomial coefficients. In particular, we will present in Section 3 four conjectural series for 1/\pi of the following new type:

    Type Ⅷ. \sum_{k = 0}^\infty \frac{a+dk}{m^k}T_k(b,c)T_{k}(b_*,c_*)^2,

    where a,b,b_*,c,c_*,d,m are integers with mbb_*cc_*d(b^2-4c)(b_*^2-4c_*)(b^2c_*-b_*^2c)\not = 0 .

    Unlike Ramanujan-type series given by others, all our series for 1/\pi of types Ⅰ-Ⅷ have the general term involving a product of three generalized central trinomial coefficients.

    Motivated by the author's effective way to find new series for 1/\pi (cf. Sun [35]), we formulate the following general characterization of rational Ramanujan-type series for 1/\pi via congruences.

    Conjecture 1.1 (General Criterion for Rational Ramanujan-type Series for 1/\pi ). Suppose that the series \sum_{k = 0}^\infty \frac{bk+c}{m^k}a_k converges, where (a_k)_{k{\geq}0} is an integer sequence and b,c,m are integers with bcm\not = 0 . Suppose also that there are no a,x \in \mathbb{Z} such that {a_n} = a(_n^{2n})\sum\nolimits_{k = 0}^n ( _k^{2k}{)^2}(_{n - k}^{2(n - k)}){x^{n - k}} for all n \in \mathbb{N}. Let r\in\{1,2,3\} and let {d_1}, \ldots ,{d_r} \in {\mathbb{Z}^ + } with \sqrt{d_i/d_j} irrational for all distinct i,j\in\{1,\ldots,r\} . Then

    \begin{equation} \sum\limits_{k = 0}^\infty \frac{bk+c}{m^k}a_k = \frac{\sum\limits_{i = 1}^r\lambda_i\sqrt{d_i}}{\pi} \end{equation} (1.98)

    for some nonzero rational numbers \lambda_1,\ldots,\lambda_r if and only if there are positive integers d_j\ (r<j \leq 3) and rational numbers c_1,c_2,c_3 with \prod_{i = 1}^rc_i\not = 0 , such that for any prime p>3 with p\nmid m\prod_{i = 1}^rd_i and c_1,c_2,c_3\in{\Bbb Z}_p we have

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{bk+c}{m^k}a_k{\equiv} p\left(\sum\limits_{i = 1}^rc_i \left( \frac{{\varepsilon}_id_i}p \right)+\sum\limits_{r < j \leq 3}c_j \left( \frac{d_j}p \right)\right) \ ({\rm{mod}}\ {p^2}), \end{equation} (1.99)

    where {\varepsilon}_i\in\{\pm1\} , {\varepsilon}_i = -1 if d_i is not an integer square, and c_2 = c_3 = 0 if r = 1 and {\varepsilon}_1 = 1 .

    For a Ramanujan-type series of the form (1.98), we call r its rank. We believe that there are some Ramanujan-type series of rank three but we have not yet found such a series.

    Conjecture 1.2. Let (a_n)_{n{\geq}0} be an integer sequence with no a,x \in \mathbb{Z} such that {a_n} = a(_n^{2n})\sum\nolimits_{k = 0}^n ( _k^{2k}{)^2}(_{n - k}^{2(n - k)}){x^{n - k}} for all n \in \mathbb{N}, and let b,c,m,d_1,d_2,d_3\in{\Bbb Z} with bcm\not = 0 . Assume that \lim_{n\to+\infty}\root n\of{|a_n|} = r<|m| , and \pi\sum_{k = 0}^\infty \frac{bk+c}{m^k}a_k is an algebraic number. Suppose that c_1,c_2,c_3 \in \mathbb{Q} with c_1+c_2+c_3 = a_0c , and

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{bk+c}{m^k}a_k{\equiv} p \left(c_1 \left( \frac{d_1}p \right)+c_2 \left( \frac{d_2}p \right)+c_3 \left( \frac{d_3}p \right) \right)\ ({\rm{mod}}\ {p^2}) \end{equation} (1.100)

    for all primes p>3 with p\nmid d_1d_2d_3m and c_1,c_2,c_3\in{\Bbb Z}_p . Then, for any prime p>3 with p\nmid m , c_1,c_2,c_3\in{\Bbb Z}_p and ( \frac{d_1}p) = ( \frac{d_2}p) = ( \frac{d_3}p) = {\delta}\in\{\pm1\} , we have

    \begin{equation*} \label{pn} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{bk+c}{m^k}a_k-p{\delta}\sum\limits_{k = 0}^{n-1} \frac{bk+c}{m^k}a_k\right)\in{\Bbb Z}_p \ \ \mathit{\text{for all}}\ n\in{\Bbb Z}^+. \end{equation*}

    Joint with the author's PhD student Chen Wang, we pose the following conjecture.

    Conjecture 1.3 (Chen Wang and Z.-W. Sun). Let (a_k)_{k{\geq}0} be an integer sequence with a_0 = 1 . Let b,c,m,d_1,d_2,d_3\in{\Bbb Z} with bm\not = 0 , and let c_1,c_2,c_3 be rational numbers. If \pi\sum_{k = 0}^\infty \frac{bk+c}{m^k}a_k is an algebraic number, and the congruence (1.100) holds for all primes p>3 with p\nmid d_1d_2d_3m and c_1,c_2,c_3\in{\Bbb Z}_p , then we must have c_1+c_2+c_3 = c.

    Remark 1.2. The author [39,Conjecture 1.1(i)] conjectured that

    \sum\limits_{k = 0}^{p-1}(8k+5)T_k^2{\equiv} 3p \left( \frac{-3}p \right)\ ({\rm{mod}}\ {p^2})

    for any prime p>3 , which was confirmed by Y.-P. Mu and Z.-W. Sun [26]. This is not a counterexample to Conjecture 1.3 since \sum_{k = 0}^\infty(8k+5)T_k^2 diverges.

    All the new series and related congruences in Sections 3-9 support Conjectures 1.1-1.3. We discover the conjectural series for 1/\pi in Sections 3-9 based on the author's previous \mathsf{Philosophy about Series for} 1/\pi } stated in [35], the PSLQ algorithm to discover integer relations (cf. [13]), and the following \mathsf{Duality Principle} based on the author's experience and intuition.

    Conjecture 1.4 (Duality Principle). Let (a_k)_{k{\geq}0} be an integer sequence such that

    \begin{equation} a_{k}{\equiv} \left( \frac dp \right)D^ka_{p-1-k}\ ({\rm{mod}}\ p) \end{equation} (1.101)

    for any prime p\nmid 6dD and k\in\{0,\ldots,p-1\} , where d and D are fixed nonzero integers. If a_0,a_1,\ldots are not all zero and m is a nonzero integer such that

    \sum\limits_{k = 0}^\infty \frac{bk+c}{m^k}a_k = \frac{\lambda_1\sqrt{d_1}+\lambda_2\sqrt{d_2}+\lambda_3\sqrt{d_3}}{\pi}

    for some b,d_1,d_2,d_3\in{\Bbb Z}^+ , c\in{\Bbb Z} and \lambda_1,\lambda_2,\lambda_3\in{\Bbb Q} , then m divides D , and

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{a_k}{m^k}{\equiv} \left( \frac dp \right)\sum\limits_{k = 0}^{p-1} \frac{a_k}{(D/m)^k}\ ({\rm{mod}}\ {p^2}) \end{equation} (1.102)

    for any prime p>3 with p\nmid dD .

    Remark 1.3 (ⅰ) For any prime p>3 with p\nmid dDm , the congruence (1.102) holds modulo p by (1.101) and Fermat's little theorem. We call \sum_{k = 0}^{p-1}a_k/( \frac Dm)^k the dual of the sum \sum_{k = 0}^{p-1}a_k/m^k .

    (ⅱ) For any b,c\in{\Bbb Z} and odd prime p\nmid b^2-4c , it is known (see, e.g., [39,Lemma 2.2]) that

    \begin{equation} T_k(b,c){\equiv} \left( \frac{b^2-4c}p \right)(b^2-4c)^kT_{p-1-k}(b,c)\ ({\rm{mod}}\ p) \end{equation} (1.103)

    for all k = 0,1,\ldots,p-1 .

    For a series \sum_{k = 0}^\infty a_k with a_0,a_1,\ldots real numbers, if \lim_{k\to+\infty}a_{k+1}/a_k = r\in(-1,1) then we say that the series converge at a geometric rate with ratio r . Except for (7.1) , all other conjectural series in Sections 3-9 converge at geometric rates and thus one can easily check them numerically via a computer.

    In Section 10, we pose two curious conjectural series for \pi involving the central trinomial coefficients.

    Lemma 2.1. Let m\not = 0 and n{\geq}0 be integers. Then

    \begin{align} \sum\limits_{k = 0}^n \frac{((64-m)k^3-32k^2-16k+8) \binom{2k}k^3}{(2k-1)^2m^k} = & \frac{8(2n+1)}{m^n} \binom{2n}n^3, \end{align} (2.1)
    \begin{align} \sum\limits_{k = 0}^n \frac{((64-m)k^3-96k^2+48k-8) \binom{2k}k^3}{(2k-1)^3m^k} = & \frac 8{m^n} \binom{2n}n^3, \end{align} (2.2)
    \begin{align} \sum\limits_{k = 0}^n \frac{((108-m)k^3-54k^2-12k+6) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)m^k} = & \frac{6(3n+1)}{m^n} \binom{2n}n^2 \binom{3n}n, \end{align} (2.3)
    \begin{equation} \begin{aligned} &\sum\limits_{k = 0}^n \frac{((108-m)k^3-(54+m)k^2-12k+6) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)m^k} \\&\qquad\qquad\qquad = \frac{6(3n+1)}{(n+1)m^n} \binom{2n}n^2 \binom{3n}n, \end{aligned} \end{equation} (2.4)
    \begin{equation} \begin{aligned} &\sum\limits_{k = 0}^n \frac{((256-m)k^3-128k^2-16k+8) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)m^k} \\&\qquad\qquad\qquad = \frac{8(4n+1)}{m^n} \binom{2n}n^2 \binom{4n}{2n}, \end{aligned} \end{equation} (2.5)
    \begin{equation} \begin{aligned} &\sum\limits_{k = 0}^n \frac{((256-m)k^3-(128+m)k^2-16k+8) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)m^k} \\&\qquad\qquad\qquad = \frac{8(4n+1)}{(n+1)m^n} \binom{2n}n^2 \binom{4n}{2n}, \end{aligned} \end{equation} (2.6)
    \begin{equation} \begin{aligned} &\sum\limits_{k = 0}^n \frac{((1728-m)k^3-864k^2-48k+24) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)m^k} \\&\qquad\qquad\qquad = \frac{24(6n+1)}{m^n} \binom{2n}n \binom{3n}n \binom{6n}{3n}, \end{aligned} \end{equation} (2.7)
    \begin{equation} \begin{aligned} &\sum\limits_{k = 0}^n \frac{((1728-m)k^3-(864+m)k^2-48k+24) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)m^k} \\&\qquad\qquad\qquad = \frac{24(6n+1)}{(n+1)m^n} \binom{2n}n \binom{3n}n \binom{6n}{3n}. \end{aligned} \end{equation} (2.8)

    Remark 2.1. The eight identities in Lemma 2.1 can be easily proved by induction on n . In light of Stirling's formula, n!\sim\sqrt{2\pi n}(n/e)^n as n\to+\infty , we have

    \begin{gather} \binom{2n}n\sim \frac{4^n}{\sqrt{n\pi}},\ \ \binom{2n}n \binom{3n}n\sim \frac{\sqrt3\, 27^n}{2n\pi}, \end{gather} (2.9)
    \begin{gather} \binom{2n}n \binom{4n}{2n}\sim \frac{64^n}{\sqrt2\, n\pi},\ \ \binom{3n}n \binom{6n}n\sim \frac{432^n}{2n\pi}. \end{gather} (2.10)

    Proof of Theorem 1.1. Just apply Lemma 2.1 and the 36 known rational Ramanujan-type series listed in [16]. Let us illustrate the proofs by showing (1.1), (1.2), (1.71) and (1.72) in details.

    By (2.1) with m = -64 , we have

    \sum\limits_{k = 0}^\infty \frac{(16k^3-4k^2-2k+1) \binom{2k}k^3}{(2k-1)^2(-64)^k} = \lim\limits_{n\to+\infty} \frac{2n+1}{(-64)^n} \binom{2n}n^3 = 0.

    Note that

    16k^3-4k^2-2k+1 = (4k+1)(2k-1)^2+2k(4k-1)

    and recall Bauer's series

    \sum\limits_{k = 0}^\infty(4k+1) \frac{ \binom{2k}k^3}{(-64)^k} = \frac2{\pi}.

    So, we get

    \sum\limits_{k = 0}^\infty \frac{k(4k-1) \binom{2k}k^3}{(2k-1)^2(-64)^k} = - \frac12\sum\limits_{k = 0}^\infty(4k+1) \frac{ \binom{2k}k^3}{(-64)^k} = - \frac1{\pi}.

    This proves (1.1). By (2.2) with m = -64 , we have

    \sum\limits_{k = 0}^n \frac{(4k-1)(4k^2-2k+1) \binom{2k}k^3}{(2k-1)^3(-64)^k} = \frac{ \binom{2n}n^3}{(-64)^n}

    and hence

    \sum\limits_{k = 0}^\infty \frac{(2k(2k-1)(4k-1)+4k-1) \binom{2k}k^3}{(2k-1)^3(-64)^k} = \lim\limits_{n\to+\infty} \frac{ \binom{2n}n^3}{(-64)^n} = 0.

    Combining this with (1.1) we immediately get (1.2) .

    In view of (2.7) with m = -640320^3 , we have

    \begin{align*} &\sum\limits_{k = 0}^n \frac{(10939058860032072k^3-36k^2-2k+1) \binom{2k}k \binom{3k}k \binom{6k}{3k}} {(2k-1)(6k-1)(-640320)^{3k}} \\ = & \frac{6n+1}{(-640320)^{3n}} \binom{2n}n \binom{3n}n \binom{6n}{3n}. \end{align*}

    and hence

    \sum\limits_{k = 0}^\infty \frac{(10939058860032072k^3-36k^2-2k+1) \binom{2k}k \binom{3k}k \binom{6k}{3k}} {(2k-1)(6k-1)(-640320)^{3k}} = 0.

    In 1987, D. V. Chudnovsky and G. V. Chudnovsky [8] got the formula

    \sum\limits_{k = 0}^\infty \frac{545140134k+13591409}{(-640320)^{3k}} \binom{2k}k \binom{3k}k \binom{6k}{3k} = \frac{3\times53360^2}{2\pi\sqrt{10005}},

    which enabled them to hold the world record for the calculation of \pi during 1989–1994. Note that

    \begin{align*} &10939058860032072k^3-36k^2-2k+1 \\ = &1672209(2k-1)(6k-1)(545140134k+13591409) \\&+426880 (16444841148 k^2 - 1709536232 k-53241371 ) \end{align*}

    and hence

    \begin{align*} &\sum\limits_{k = 0}^\infty \frac{(16444841148 k^2 - 1709536232 k-53241371 ) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(-640320)^{3k}} \\ = &- \frac{1672209}{426880}\times \frac{3\times53360^2}{2\pi\sqrt{10005}} = -1672209 \frac{\sqrt{10005}}{\pi}. \end{align*}

    This proves (1.71) .

    By (2.8) with m = -640320^3 , we have

    \begin{align*} &\sum\limits_{k = 0}^n \frac{(10939058860032072k^3 +10939058860031964k^2-2k+1) \binom{2k}k \binom{3k}k \binom{6k}{3k}} {(k+1)(2k-1)(6k-1)(-640320)^{3k}} \\&\qquad\qquad = \frac{6n+1}{(n+1)(-640320)^{3n}} \binom{2n}n \binom{3n}n \binom{6n}{3n} \end{align*}

    and hence

    \sum\limits_{k = 0}^\infty \frac{(10939058860032072k^3 +10939058860031964k^2-2k+1) \binom{2k}k \binom{3k}k \binom{6k}{3k}} {(k+1)(2k-1)(6k-1)(-640320)^{3k}} = 0.

    Note that

    \begin{align*} &2802461(10939058860032072k^3 +10939058860031964k^2-2k+1) \\ = &1864188626454(k+1)(16444841148 k^2 - 1709536232 k-53241371)+5P(k). \end{align*}

    Therefore, with the help of (1.71) we get

    \begin{align*} &\sum\limits_{k = 0}^\infty \frac{P(k) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(-640320)^{3k}} \\ = &- \frac{1864188626454}5\times(-1672209) \frac{\sqrt{10005}}{\pi} = 18\times557403^3 \frac{\sqrt{10005}}{5\pi}. \end{align*}

    This proves (1.72) .

    The identities (1.3)–(1.70) can be proved similarly.

    Lemma 2.2. Let m and n>0 be integers. Then

    \begin{align} \sum\limits_{k = 1}^n \frac{m^k((m-64)k^3-32k^2+16k+8)}{(2k+1)^2k^3 \binom{2k}k^3} = & \frac{m^{n+1}}{(2n+1)^2 \binom{2n}n^3}-m, \end{align} (2.11)
    \begin{align} \sum\limits_{k = 1}^n \frac{m^k((m-64)k^3-96k^2-48k-8)}{(2k+1)^3k^3 \binom{2k}k^3} = & \frac{m^{n+1}}{(2n+1)^3 \binom{2n}n^3}-m, \end{align} (2.12)
    \begin{align} \sum\limits_{k = 1}^n \frac{m^k((m-108)k^3-54k^2+12k+6)}{(2k+1)(3k+1)k^3 \binom{2k}k^2 \binom{3k}k} = & \frac{m^{n+1}}{(2n+1)(3n+1) \binom{2n}n^2 \binom{3n}n}-m, \end{align} (2.13)
    \begin{equation} \begin{aligned} &\sum\limits_{1 < k \leq n} \frac{m^k((m-108)k^3-(54+m)k^2+12k+6)}{(k-1)(2k+1)(3k+1)k^3 \binom{2k}k^2 \binom{3k}k} \\&\qquad\qquad = \frac{m^{n+1}}{n(2n+1)(3n+1) \binom{2n}n^2 \binom{3n}n}- \frac{m^2}{144}, \end{aligned} \end{equation} (2.14)
    \begin{equation} \begin{aligned} &\sum\limits_{k = 1}^n \frac{m^k((m-256)k^3-128k^2+16k+8)}{(2k+1)(4k+1)k^3 \binom{2k}k^2 \binom{4k}{2k}} \\&\qquad\qquad = \frac{m^{n+1}}{(2n+1)(4n+1) \binom{2n}n^2 \binom{4n}{2n}}-m, \end{aligned} \end{equation} (2.15)
    \begin{equation} \begin{aligned} &\sum\limits_{1 < k \leq n} \frac{m^k((m-256)k^3-(128+m)k^2+16k+8)}{(k-1)(2k+1)(4k+1)k^3 \binom{2k}k^2 \binom{4k}{2k}} \\&\qquad\qquad\quad = \frac{m^{n+1}}{n(2n+1)(4n+1) \binom{2n}n^2 \binom{4n}{2n}}- \frac{m^2}{360}. \end{aligned} \end{equation} (2.16)

    Remark 2.2. This can be easily proved by induction on n .

    Proof of Theorem 1.2. We just apply Lemma 2.2 and use the known identities:

    \begin{gather*} \sum\limits_{k = 1}^\infty \frac{21k-8}{k^3 \binom{2k}k^3} = \frac{\pi^2}6, \ \ \sum\limits_{k = 1}^\infty \frac{(4k-1)(-64)^k}{k^3 \binom{2k}k^3} = -16G, \\\sum\limits_{k = 1}^\infty \frac{(3k-1)(-8)^k}{k^3 \binom{2k}k^3} = -2G, \ \ \sum\limits_{k = 1}^\infty \frac{(3k-1)16^k}{k^3 \binom{2k}k^3} = \frac{\pi^2}2, \\\sum\limits_{k = 1}^\infty \frac{(15k-4)(-27)^{k-1}}{k^3 \binom{2k}k^2 \binom{3k}k} = K,\ \ \sum\limits_{k = 1}^\infty \frac{(5k-1)(-144)^k}{k^3 \binom{2k}k^2 \binom{4k}{2k}} = - \frac{45}2K, \\\sum\limits_{k = 1}^\infty \frac{(11k-3)64^k}{k^2 \binom{2k}k^2 \binom{3k}k} = 8\pi^2,\ \sum\limits_{k = 1}^\infty \frac{(10k-3)8^k}{k^3 \binom{2k}k^2 \binom{3k}k} = \frac{\pi^2}2,\ \ \sum\limits_{k = 1}^\infty \frac{(35k-8)81^k}{k^3 \binom{2k}k^2 \binom{4k}{2k}} = 12\pi^2. \end{gather*}

    Here, the first identity was found and proved by D. Zeilberger [52] in 1993. The second, third and fourth identities were obtained by J. Guillera [17] in 2008. The fifth identity on K was conjectured by Sun [33] and later confirmed by K. Hessami Pilehrood and T. Hessami Pilehrood [22] in 2012. The last four identities were also conjectured by Sun [33], and they were later proved in the paper [18,Theorem 3] by Guillera and M. Rogers.

    Let us illustrate our proofs by proving (1.77)-(1.79) and (1.82)-(1.83) in details.

    In view of (2.11) with m = 16 , we have

    \sum\limits_{k = 1}^n \frac{16^k(-48k^3-32k^2+16k+8)}{(2k+1)^2k^3 \binom{2k}k^3} = \frac{16^{n+1}}{(2n+1)^2 \binom{2n}n^3}-16

    for all n\in{\Bbb Z}^+ , and hence

    \sum\limits_{k = 1}^\infty \frac{16^k(6k^3+4k^2-2k-1)}{(2k+1)^2k^3 \binom{2k}k^3} = \lim\limits_{n\to+\infty} \left( \frac{-2\times16^n}{(2n+1)^2 \binom{2n}n^3}+2 \right) = 2.

    Notice that

    2(6k^3+4k^2-2k-1) = (2k+1)^2(3k-1)-(3k+1).

    So we have

    -\sum\limits_{k = 1}^\infty \frac{(3k+1)16^k}{(2k+1)^2k^3 \binom{2k}k^3} = 2\times2-\sum\limits_{k = 1}^\infty \frac{(3k-1)16^k}{k^3 \binom{2k}k^3} = 4- \frac{\pi^2}2

    and hence (1.77) holds.

    By (2.11) with m = -64 , we have

    \sum\limits_{k = 1}^n \frac{(-64)^k(-128k^3-32k^2+16k+8)}{(2k+1)^2k^3 \binom{2k}k^3} = \frac{(-64)^{n+1}}{(2n+1)^2 \binom{2n}n^3}+64

    for all n\in{\Bbb Z}^+ , and hence

    \sum\limits_{k = 1}^\infty \frac{(-64)^k(16k^3+4k^2-2k-1)}{(2k+1)^2k^3 \binom{2k}k^3} = -8+\lim\limits_{n\to+\infty} \frac{8(-64)^n}{(2n+1)^2 \binom{2n}n^3} = -8.

    Since 16k^3+4k^2-2k-1 = (4k-1)(2k+1)^2-2k(4k+1) and

    \sum\limits_{k = 1}^\infty \frac{(4k-1)(-64)^k}{k^3 \binom{2k}k^3} = -16G,

    we see that

    -16G-2\sum\limits_{k = 1}^\infty \frac{(4k+1)(-64)^k}{(2k+1)^2k^2 \binom{2k}k^3} = -8

    and hence (1.78) holds. In light of (2.12) with m = -64 , we have

    \sum\limits_{k = 1}^n \frac{(-64)^k(-128k^3-96k^2-48k-8)}{(2k+1)^3k^3 \binom{2k}k^3} = \frac{(-64)^{n+1}}{(2n+1)^3 \binom{2n}n^3}+64

    for all n\in{\Bbb Z}^+ , and hence

    \sum\limits_{k = 1}^\infty \frac{(-64)^k(16k^3+12k^2+6k+1)}{(2k+1)^3k^3 \binom{2k}k^3} = -8+\lim\limits_{n\to+\infty} \frac{8(-64)^n}{(2n+1)^3 \binom{2n}n^3} = -8.

    Since 16k^3+12k^2+6k+1 = 2k(2k+1)(4k+1)+(4k+1) , with the aid of (1.78) we obtain

    \sum\limits_{k = 1}^\infty \frac{(4k+1)(-64)^k}{(2k+1)^3k^3 \binom{2k}k^3} = -8-2(4-8G) = 16G-16.

    This proves (1.79).

    By (2.13) with m = -27 , we have

    \sum\limits_{k = 1}^\infty \frac{(45k^3+18k^2-4k-2)(-27)^k}{(2k+1)(3k+1)k^3 \binom{2k}k^2 \binom{3k}k} = -9.

    As

    2(45k^3+18k^2-4k-2) = (15k-4)(2k+1)(3k+1)-3k(5k+1)

    and

    \sum\limits_{k = 1}^\infty \frac{(15k-4)(-27)^k}{k^3 \binom{2k}k^2 \binom{3k}k} = -27K,

    we see that (1.82) follows. By (2.14) with m = -27 , we have

    -3\sum\limits_{k = 2}^\infty \frac{(-27)^k(45k^3+9k^2-4k-2)}{(k-1)(2k+1)(3k+1)k^3 \binom{2k}k^2 \binom{3k}k} = - \frac{(-27)^2}{144}

    and hence

    \sum\limits_{k = 2}^\infty \frac{(-27)^k(45k^3+9k^2-4k-2)}{(k-1)(2k+1)(3k+1)k^3 \binom{2k}k^2 \binom{3k}k} = \frac{27}{16}.

    As

    45k^3+9k^2-4k-2 = 9(k-1)k(5k+1)+(45k^2+5k-2),

    with the aid of (1.82) we get

    \begin{align*} &\sum\limits_{k = 2}^\infty \frac{(-27)^k(45k^2+5k-2)}{(k-1)(2k+1)(3k+1)k^3 \binom{2k}k^2 \binom{3k}k} \\ = & \frac{27}{16}-9 \left(6-9K- \frac{6(-27)}{12^2} \right) = \frac{27}{16}(48K-37) \end{align*}

    and hence (1.83) follows.

    Other identities in Theorem 1.2 can be proved similarly.

    For integers n{\geq} k{\geq}0 , we define

    \begin{equation} s_{n,k}: = \frac1{ \binom nk}\sum\limits_{i = 0}^{k} \binom n{2i} \binom n{2(k-i)} \binom{2i}i \binom{2(k-i)}{k-i}. \end{equation} (2.17)

    For n\in{\Bbb N} we set

    \begin{equation} t_n: = \sum\limits_{0 < k \leq n} \binom{n-1}{k-1}(-1)^k4^{n-k}s_{n+k,k}. \end{equation} (2.18)

    Lemma 2.3. For any n\in{\Bbb N} , we have

    \begin{equation} \sum\limits_{k = 0}^n \binom nk(-1)^k4^{n-k}s_{n+k,k} = f_n \end{equation} (2.19)

    and

    \begin{equation} (2n+1)t_{n+1}+8nt_n = (2n+1)f_{n+1}-4(n+1)f_n, \end{equation} (2.20)

    where f_n denotes the Franel number \sum_{k = 0}^n \binom nk^3 .

    Proof. For n,i,k\in{\Bbb N} with i \leq k , we set

    F(n,i,k) = {n \choose k} \frac{(-1)^k 4^{n-k}}{ \binom{n+k}k} {n+k \choose 2i}{2i \choose i} {n+k \choose 2(k-i)}{2(k-i) \choose k-i}.

    By the telescoping method for double summation [7], for

    {\mathcal F}(n,i,k): = F(n,i,k) + \frac{7n^2+21n+16}{8(n+1)^2} F(n+1,i,k) - \frac{(n+2)^2}{8(n+1)^2} F(n+2,i,k)

    with 0 \leq i \leq k , we find that

    {\mathcal F}(n,i,k) = (G_1(n,i+1,k)-G_1(n,i,k)) + (G_2(n,i,k+1)-G_2(n,i,k)),

    where

    G_1(n,i,k): = \frac{i^2(-k+i-1)(-1)^{k+1} 4^{n-k} n!^2 (n+k)! p(n,i,k)}{(2n+3)(n-k+2)!(n+k+2-2i)!(n-k+2i)!(i!(k-i+1)!)^2}

    and

    G_2(n,i,k): = \frac{ 2(k-i)(-1)^{k} 4^{n-k} n!^2 (n+k)! q(n,i,k)} {(2n+3) (n-k+2)! (n+k-2i+1)! (n-k+2i+2)! (i!(k-i)!)^2},

    with (-1)!,(-2)!,\ldots regarded as +\infty , and p(n,i,k) and q(n,i,k) given by

    \begin{align*} &-10n^4+(i-10k-68)n^3+(-24i^2+(32k+31)i+2k^2-67k-172)n^2\\ &+(36i^3+(-68k-124)i^2+(39k^2+149k+104)i+2k^3-8k^2-145k-192)n\\ &+60i^3+(-114k-140)i^2+(66k^2+160k+92)i+3k^3-19k^2-102k-80 \end{align*}

    and

    \begin{align*} &10(i-k)n^4+(-20i^2+(46k+47)i-6k^2-47k)n^3\\ +&(72i^3+(-60k-38)i^2+(22k^2+145k+90)i+4k^3-11k^2-90k)n^2\\ +&(72k+156)i^3n+(-72k^2-60k-10)i^2n+(18k^3+4k^2+165k+85)in \\+&(22k^3-5k^2-85k)n +(120k+60)i^3+(-120k^2+68k-4)i^2 \\+&(30k^3-56k^2+86k+32)i+26k^3-6k^2-32k \end{align*}

    respectively. Therefore

    \begin{align*} &\sum\limits_{k = 0}^{n+2} \sum\limits_{i = 0}^k {\mathcal F}(n,i,k) \\ = & \sum\limits_{k = 0}^{n+2} (G_1(n,k+1,k)-G_1(n,0,k)) + \sum\limits_{i = 0}^{n+2} (G_2(n,i,n+3)-G_2(n,i,i)) \\ = &\sum\limits_{k = 0}^{n+2}(0-0)+\sum\limits_{i = 0}^{n+2}(0-0) = 0, \end{align*}

    and hence

    u(n): = \sum\limits_{k = 0}^n {n \choose k} (-1)^k 4^{n-k} s_{n+k,k}

    satisfies the recurrence relation

    8(n+1)^2 u(n) + (7n^2+21n+16) u(n+1) - (n+2)^2 u(n+2) = 0.

    As pointed out by J. Franel [14], the Franel numbers satisfy the same recurrence. Note also that u(0) = f_0 = 1 and u(1) = f_1 = 2 . So we always have u(n) = f_n . This proves (2.19).

    The identity (2.20) can be proved similarly. In fact, if we use v(n) denote the left-hand side or the right-hand side of (2.20), then we have the recurrence

    \begin{align*} &8(n+1)(n+2)(18n^3+117n^2+249n+172)v(n) \\&+(126n^5+1197n^4+4452n^3+8131n^2+7350n+2656)v(n+1) \\ = &(n+3)^2(18n^3+63n^2+69n+22)v(n+2). \end{align*}

    In view of the above, we have completed the proof of Lemma 2.3.

    Lemma 2.4. For any c\in{\Bbb Z} and n\in{\Bbb N} , we have

    \begin{equation} S_n(4,c) = \sum\limits_{k = 0}^{\lfloor n/2\rfloor} \binom{n-k}k \binom{2(n-k)}{n-k}c^k4^{n-2k}s_{n,k}. \end{equation} (2.21)

    Proof. For each k = 0,\ldots,n , we have

    \begin{align*} T_k(4,c)T_{n-k}(4,c) = &\sum\limits_{i = 0}^{\lfloor k/2\rfloor} \binom {k}{2i} \binom{2i}i4^{k-2i}c^i \sum\limits_{j = 0}^{\lfloor(n-k)/2\rfloor} \binom{n-k}{2j} \binom{2j}j4^{n-k-2j}c^j \\ = &\sum\limits_{r = 0}^{\lfloor n/2\rfloor}c^r4^{n-2r}\sum\limits_{i,j\in{\Bbb N}\atop i+j = r} \binom k{2i} \binom{n-k}{2j} \binom{2i}i \binom{2j}j. \end{align*}

    If i,j\in{\Bbb N} and i+j = r \leq n/2 , then

    \begin{align*} \sum\limits_{k = 0}^n \binom nk^2 \binom{k}{2i} \binom{n-k}{2j} = & \binom n{2i} \binom n{2j}\sum\limits_{k = 2i}^{n-2j} \binom{n-2i}{k-2i} \binom{n-2j}{n-k-2j} \\ = & \binom n{2i} \binom n{2j} \binom{2n-2(i+j)}{n-2(i+j)} = \binom{2n-2r}n \binom n{2i} \binom n{2j} \end{align*}

    with the aid of the Chu-Vandermonde identity. Therefore

    \begin{align*} S_n(4,c) = &\sum\limits_{k = 0}^{\lfloor n/2\rfloor}c^k4^{n-2k} \binom{2n-2k}{n} \binom nks_{n,k} \\ = &\sum\limits_{k = 0}^{\lfloor n/2\rfloor}c^k4^{n-2k} \binom{2n-2k}{n-k} \binom{n-k}ks_{n,k}. \end{align*}

    This proves (2.21).

    Lemma 2.5. For k\in{\Bbb N} and l\in{\Bbb Z}^+ , we have

    \begin{equation} s_{k+l,k} \leq (2k+1)4^kl \binom{k+l}l. \end{equation} (2.22)

    Proof. Let n = k+l . Then

    \begin{align*} \binom nks_{n,k} \leq&\sum\limits_{i,j\in{\Bbb N}\atop i+j = k} \binom n{2i} \binom n{2j}\sum\limits_{i,j\in{\Bbb N}\atop i+j = k} \binom{2i}i \binom{2j}j \\ \leq&\sum\limits_{s,t\in{\Bbb N}\atop s+t = 2k} \binom ns \binom nt\sum\limits_{i,j\in{\Bbb N}\atop i+j = k}4^i4^j = \binom{2n}{2k}(k+1)4^k \end{align*}

    and

    \begin{align*} \frac{ \binom{2n}{2k}}{ \binom nk} = & \frac{ \binom{2n}{2l}}{ \binom nl} = \prod\limits_{j = 0}^{l-1} \frac{2(j+k)+1}{2j+1} \\ \leq&(2k+1)\prod\limits_{0 < j < l} \frac{2(j+k)}{2j} \\ = &(2k+1) \binom{k+l-1}{l-1}. \end{align*}

    Hence

    s_{k+l,k} \leq (k+1)4^k(2k+1) \frac{l}{k+l} \binom{k+l}l \leq (2k+1)4^kl \binom{k+l}l.

    This proves (2.22).

    To prove Theorem 1.3, we need an auxiliary theorem.

    Theorem 2.6. Let a and b be real numbers. For any integer m with |m|{\geq}94 , we have

    \begin{equation} \sum\limits_{n = 0}^\infty(an+b) \frac{S_n(4,-m)}{m^n} = \frac1{m+16}\sum\limits_{n = 0}^\infty(2a(m+4)n-8a+b(m+16)) \frac{ \binom{2n}nf_n}{m^n}. \end{equation} (2.23)

    Proof. Let N\in{\Bbb N} . In view of (2.21),

    \begin{align*} \sum\limits_{n = 0}^N \frac{S_n(4,-m)}{m^n} = &\sum\limits_{n = 0}^N \frac1{m^n}\sum\limits_{k = 0}^{\lfloor n/2\rfloor}(-m)^k4^{n-2k} \binom{2n-2k}{n-k} \binom{n-k}ks_{n,k} \end{align*}
    \begin{align*} = &\sum\limits_{l = 0}^N \frac{ \binom{2l}l}{m^l}\sum\limits_{k = 0}^{N-l} \binom lk(-1)^k4^{l-k}s_{l+k,k} \\ = &\sum\limits_{l = 0}^{\lfloor N/2\rfloor} \frac{ \binom{2l}l}{m^l}\sum\limits_{k = 0}^l \binom lk(-1)^k4^{l-k}s_{l+k,k} \\&+\sum\limits_{N/2 < l \leq N} \frac{ \binom{2l}l}{m^l}\sum\limits_{k = 0}^{N-l} \binom lk(-1)^k4^{l-k}s_{l+k,k} \end{align*}

    and similarly

    \begin{align*} \sum\limits_{n = 0}^N \frac{nS_n(4,-m)}{m^n} = &\sum\limits_{l = 0}^N \frac{ \binom{2l}l}{m^l}\sum\limits_{k = 0}^{N-l} \binom lk(-1)^k4^{l-k}(k+l)s_{l+k,k} \\ = &\sum\limits_{l = 0}^N \frac{l \binom{2l}l}{m^l}\sum\limits_{k = 0}^{N-l} \left( \binom lk+ \binom{l-1}{k-1} \right)(-1)^k4^{l-k}s_{l+k,k} \\ = &\sum\limits_{l = 0}^{\lfloor N/2\rfloor} \frac{l \binom{2l}l}{m^l}\sum\limits_{k = 0}^{l} \left( \binom lk+ \binom{l-1}{k-1} \right)(-1)^k4^{l-k}s_{l+k,k} \\&+\sum\limits_{N/2 < l \leq N} \frac{l \binom{2l}l}{m^l}\sum\limits_{k = 0}^{N-l} \left( \binom lk+ \binom{l-1}{k-1} \right)(-1)^k4^{l-k}s_{l+k,k}, \end{align*}

    where we consider \binom x{-1} as 0 .

    If l is an integer in the interval (N/2,N] , then by Lemma 2.5 we have

    \begin{align*} & \bigg|\sum\limits_{k = 0}^{N-l} \binom lk(-1)^k4^{l-k}s_{l+k,k} \bigg| \\ \leq&\sum\limits_{k = 0}^l \binom lk4^{l-k}s_{l+k,k} \leq\sum\limits_{k = 0}^l \binom lk4^{l-k}(2k+1)4^kl \binom{k+l}l \\ \leq& l(2l+1)4^l\sum\limits_{k = 0}^l \binom lk \binom{l+k}k = l(2l+1)4^lP_l(3), \end{align*}

    where P_l(x) is the Legendre polynomial of degree l . Thus

    \begin{align*} & \bigg|\sum\limits_{N/2 < l \leq N} \frac{ \binom{2l}l}{m^l}\sum\limits_{k = 0}^{N-l} \binom lk(-1)^k4^{l-k}s_{l+k,k} \bigg| \\ \leq&\sum\limits_{N/2 < l \leq N}l(2l+1) \left( \frac{16}m \right)^lP_l(3) \leq\sum\limits_{l > N/2}l(2l+1)P_l(3) \left( \frac{16}m \right)^l \end{align*}

    and

    \begin{align*} & \bigg|\sum\limits_{N/2 < l \leq N} \frac{l \binom{2l}l}{m^l}\sum\limits_{k = 0}^{N-l} \left( \binom lk+ \binom{l-1}{k-1} \right)(-1)^k4^{l-k}s_{l+k,k} \bigg| \\ \leq&\sum\limits_{N/2 < l \leq N} \frac{l4^l}{m^l}\sum\limits_{k = 0}^l 2 \binom lk4^{l-k}s_{l+k,k} \end{align*}
    \begin{align*} \leq&\sum\limits_{N/2 < l \leq N}2l^2(2l+1) \left( \frac{16}m \right)^lP_l(3) \\ \leq&2\sum\limits_{l > N/2}l^2(2l+1)P_l(3) \left( \frac{16}m \right)^l. \end{align*}

    Recall that

    P_l(3) = T_l(3,2)\sim \frac{(3+2\sqrt2)^{l+1/2}}{2\root4\of{2}\sqrt{l\pi}}\ \ \text{as}\ l\to+\infty.

    As |m|{\geq}94 , we have |m|>16(3+2\sqrt2)\approx 93.255 and hence

    \sum\limits_{l = 0}^\infty l^2(2l+1)P_l(3) \left( \frac{16}m \right)^l

    converges. Thus

    \lim\limits_{N\to+\infty}\sum\limits_{l > N/2}l(2l+1)P_l(3) \left( \frac{16}m \right)^l = 0 = \lim\limits_{N\to+\infty}\sum\limits_{l > N/2}l^2(2l+1)P_l(3) \left( \frac{16}m \right)^l

    and hence by the above we have

    \sum\limits_{n = 0}^\infty \frac{S_n(4,-m)}{m^n} = \sum\limits_{l = 0}^{\infty} \frac{ \binom{2l}l}{m^l}\sum\limits_{k = 0}^l \binom lk(-1)^k4^{l-k}s_{l+k,k}

    and

    \sum\limits_{n = 0}^\infty \frac{nS_n(4,-m)}{m^n} = \sum\limits_{l = 0}^{\infty} \frac{l \binom{2l}l}{m^l}\sum\limits_{k = 0}^{l} \left( \binom lk+ \binom{l-1}{k-1} \right)(-1)^k4^{l-k}s_{l+k,k}.

    Therefore, with the aid of (2.19), we obtain

    \begin{equation} \sum\limits_{n = 0}^\infty \frac{S_n(4,-m)}{m^n} = \sum\limits_{n = 0}^\infty \frac{ \binom{2n}n}{m^n}f_n \end{equation} (2.24)

    and

    \begin{equation} \sum\limits_{n = 0}^\infty \frac{nS_n(4,-m)}{m^n} = \sum\limits_{n = 0}^\infty \frac{n \binom{2n}n}{m^n}(f_n+t_n). \end{equation} (2.25)

    In view of (2.25) and (2.20),

    \begin{align*} &(m+16)\sum\limits_{n = 0}^\infty \frac{nS_n(4,-m)}{m^n} \\ = &\sum\limits_{n = 1}^\infty \frac{n \binom{2n}n}{m^{n-1}}(f_n+t_n)+16\sum\limits_{n = 0}^\infty \frac{n \binom{2n}n}{m^n}(f_n+t_n) \\ = &\sum\limits_{n = 0}^\infty \frac{(n+1) \binom{2n+2}{n+1}(f_{n+1}+t_{n+1})+16n \binom{2n}n(f_n+t_n)}{m^n} \\ = &2\sum\limits_{n = 0}^\infty \frac{ \binom{2n}n}{m^n} \left((2n+1)(f_{n+1}+t_{n+1})+8n(f_n+t_n) \right) \end{align*}
    \begin{align*} = &2\sum\limits_{n = 0}^\infty \frac{ \binom{2n}n}{m^n} \left(2(2n+1)f_{n+1}+4(n-1)f_n \right) \\ = &2\sum\limits_{n = 0}^\infty \frac{(n+1) \binom{2n+2}{n+1}f_{n+1}}{m^n}+8\sum\limits_{n = 0}^\infty \frac{(n-1) \binom{2n}nf_n}{m^n} \\ = &2\sum\limits_{n = 0}^\infty \frac{n \binom{2n}nf_n}{m^{n-1}}+8\sum\limits_{n = 0}^\infty \frac{(n-1) \binom{2n}nf_n}{m^n} = 2\sum\limits_{n = 0}^\infty((m+4)n-4) \frac{ \binom{2n}nf_n}{m^n}. \end{align*}

    Combining this with (2.24), we immediately obtain the desired (2.23).

    Proof of Theorem 1.3. Let a,b,m\in{\Bbb Z} with |m|{\geq}6 . Since

    4^nT_n(1,m) = \sum\limits_{k = 0}^{\lfloor n/2\rfloor} \binom n{2k} \binom{2k}k4^{n-2k}(16m)^k = T_n(4,16m)

    for any n\in{\Bbb N} , we have 4^nS_n(1,m) = S_n(4,16m) for all n\in{\Bbb N} . Thus, in light of Theorem 2.6,

    \begin{align*} &\sum\limits_{n = 0}^\infty(an+b) \frac{S_n(1,m)}{(-4m)^n} \\ = &\sum\limits_{n = 0}^\infty(an+b) \frac{S_n(4,16m)}{(-16m)^n} \\ = & \frac1{16-16m}\sum\limits_{n = 0}^\infty(2a(4-16m)n-8a+(16-16m)b) \frac{ \binom{2n}nf_n}{(-16m)^n} \\ = & \frac1{2(m-1)}\sum\limits_{n = 0}^\infty(a(4m-1)n+a+2b(m-1)) \frac{ \binom{2n}nf_n}{(-16m)^n}. \end{align*}

    Therefore

    \begin{align*} \sum\limits_{k = 0}^\infty \frac{7k+3}{24^k}S_k(1,-6) = & \frac52\sum\limits_{k = 0}^\infty \frac{5k+1}{96^k} \binom{2k}kf_k, \\\sum\limits_{k = 0}^\infty \frac{12k+5}{(-28)^k}S_k(1,7) = &3\sum\limits_{k = 0}^\infty \frac{9k+2}{(-112)^k} \binom{2k}kf_k,\\ \sum\limits_{k = 0}^\infty \frac{84k+29}{80^k}S_k(1,-20) = &27\sum\limits_{k = 0}^\infty \frac{6k+1}{320^k} \binom{2k}kf_k, \\\sum\limits_{k = 0}^\infty \frac{3k+1}{(-100)^k}S_k(1,25) = & \frac1{16}\sum\limits_{k = 0}^\infty \frac{99k+17}{(-400)^k} \binom{2k}kf_k, \\\sum\limits_{k = 0}^\infty \frac{228k+67}{224^k}S_k(1,-56) = &5\sum\limits_{k = 0}^\infty \frac{90k+13}{896^k} \binom{2k}kf_k, \\\sum\limits_{k = 0}^\infty \frac{399k+101}{(-676)^k}S_k(1,169) = & \frac{15}{16}\sum\limits_{k = 0}^\infty \frac{855k+109}{(-2704)^k} \binom{2k}kf_k, \\ \sum\limits_{k = 0}^\infty \frac{2604k+563}{2600^k}S_k(1,-650) = &51\sum\limits_{k = 0}^\infty \frac{102k+11}{10400^k} \binom{2k}kf_k,\\ \sum\limits_{k = 0}^\infty \frac{39468k+7817}{(-6076)^k}S_k(1,1519) = &135\sum\limits_{k = 0}^\infty \frac{585k+58}{(-24304)^k} \binom{2k}kf_k, \\\sum\limits_{k = 0}^\infty \frac{41667k+7879}{9800^k}S_k(1,-2450) = & \frac{297}2\sum\limits_{k = 0}^\infty \frac{561k+53}{39200^k} \binom{2k}kf_k, \\\sum\limits_{k = 0}^\infty \frac{74613k+10711}{(-530^2)^k}S_k(1,265^2) = & \frac{23}{32}\sum\limits_{k = 0}^\infty \frac{207621k+14903}{(-1060^2)^k} \binom{2k}kf_k. \end{align*}

    It is known (cf. [5,4]) that

    \begin{gather*} \sum\limits_{k = 0}^\infty \frac{5k+1}{96^k} \binom{2k}kf_k = \frac{3\sqrt2}{\pi}, \ \ \sum\limits_{k = 0}^\infty \frac{9k+2}{(-112)^k} \binom{2k}kf_k = \frac{2\sqrt7}{\pi}, \\ \sum\limits_{k = 0}^\infty \frac{6k+1}{320^k} \binom{2k}kf_k = \frac{8\sqrt{15}}{9\pi},\ \ \sum\limits_{k = 0}^\infty \frac{99k+17}{(-400)^k} \binom{2k}kf_k = \frac{50}{\pi}, \\\sum\limits_{k = 0}^\infty \frac{90k+13}{896^k} \binom{2k}kf_k = \frac{16\sqrt7}{\pi},\ \ \sum\limits_{k = 0}^\infty \frac{855k+109}{(-2704)^k} \binom{2k}kf_k = \frac{338}{\pi}, \\\sum\limits_{k = 0}^\infty \frac{102k+11}{10400^k} \binom{2k}kf_k = \frac{50\sqrt{39}}{9\pi},\ \ \sum\limits_{k = 0}^\infty \frac{585k+58}{(-24304)^k} \binom{2k}kf_k = \frac{98\sqrt{31}}{3\pi}, \\\sum\limits_{k = 0}^\infty \frac{561k+53}{39200^k} \binom{2k}kf_k = \frac{1225\sqrt6}{18\pi}, \ \ \sum\limits_{k = 0}^\infty \frac{207621k+14903}{(-1060^2)^k} \binom{2k}kf_k = \frac{140450}{3\pi}. \end{gather*}

    So we get the identities (1.88)-(1.97) finally.

    Now we pose a conjecture related to the series (Ⅰ1)-(Ⅰ4) of Sun [34,40].

    Conjecture 3.1. We have the following identities:

    \sum\limits_{k = 0}^\infty \frac{50k+1}{(-256)^k} \binom{2k}k \binom{2k}{k+1}T_k(1,16) = \frac{8}{3\pi},\;\;\;\;\;\;\;\;(\mathrm{Ⅰ}1')
    \sum\limits_{k = 0}^\infty \frac{(100k^2-4k-7) \binom{2k}k^2T_k(1,16)}{(2k-1)^2(-256)^k} = - \frac{24}{\pi},\;\;\;\;\;\;\;\;(\mathrm{Ⅰ}1'')
    \sum\limits_{k = 0}^\infty \frac{30k+23}{(-1024)^k} \binom{2k}k \binom{2k}{k+1}T_k(34,1) = - \frac{20}{3\pi},\;\;\;\;\;\;\;(\mathrm{Ⅰ}2')
    \sum\limits_{k = 0}^\infty \frac{(36k^2-12k+1) \binom{2k}k^2T_k(34,1)}{(2k-1)^2(-1024)^k} = - \frac{6}{\pi},\;\;\;\;\;\;\;\;(\mathrm{Ⅰ}2'')
    \sum\limits_{k = 0}^\infty \frac{110k+103}{4096^k} \binom{2k}k \binom{2k}{k+1}T_k(194,1) = \frac{304}{\pi},\;\;\;\;\;\;\;\;\;(\mathrm{Ⅰ}3')
    \sum\limits_{k = 0}^\infty \frac{(20k^2+28k-11) \binom{2k}k^2T_k(194,1)}{(2k-1)^2 4096^k} = - \frac{6}{\pi},\;\;\;\;\;\;\;\;\;(\mathrm{Ⅰ}3'')
    \sum\limits_{k = 0}^\infty \frac{238k+263}{4096^k} \binom{2k}k \binom{2k}{k+1}T_k(62,1) = \frac{112\sqrt3}{3\pi},\;\;\;\;\;\;\;\;\;(\mathrm{Ⅰ}4')
    \sum\limits_{k = 0}^\infty \frac{(44k^2+4k-5) \binom{2k}k^2T_k(62,1)}{(2k-1)^2 4096^k} = - \frac{4\sqrt3}{\pi},\;\;\;\;\;\;\;\;\;(\mathrm{Ⅰ}4'')
    \sum\limits_{k = 0}^\infty \frac{6k+1}{256^k} \binom{2k}k^2T_k(8,-2) = \frac{2}{\pi}\sqrt{8+6\sqrt2},\;\;\;\;\;\;\;\;\;(\mathrm{Ⅰ}5)
    \sum\limits_{k = 0}^\infty \frac{2k+3}{256^k} \binom{2k}k \binom{2k}{k+1}T_k(8,-2) = \frac{6\sqrt{8+6\sqrt2}-16\root4\of{2}}{3\pi},\;\;\;\;\;\;\;\;\;(\mathrm{Ⅰ}5')
    \sum\limits_{k = 0}^\infty \frac{(4k^2+2k-1) \binom{2k}k^2T_k(8,-2)}{(2k-1)^2 256^k} = - \frac{3\root4\of{2}}{4\pi}.\;\;\;\;\;\;\;\;\;(\mathrm{Ⅰ}5'')

    Remark 3.1. For each k\in{\Bbb N} , we have

    ((1+\lambda_0-\lambda_1)k+\lambda_0)C_k = (k+\lambda_0) \binom{2k}k-(k+\lambda_1) \left(\begin{array}{c}2 k \\ k+1\end{array}\right)

    since \binom{2k}k = (k+1)C_k and \binom{2k}{k+1} = kC_k . Thus, for example, [40,(I1)] and (I1 ' ) together imply that

    \sum\limits_{k = 0}^\infty \frac{26k+5}{(-256)^k} \binom{2k}kC_kT_k(1,16) = \frac{16}{\pi},

    and (I5) and (I5 ' ) imply that

    \sum\limits_{k = 0}^\infty \frac{2k-1}{256^k} \binom{2k}kC_kT_k(8,-2) = \frac4{\pi} \left(\sqrt{8+6\sqrt2}-4\root4\of2 \right).

    For the conjectural identities in Conjecture 3.1, we have conjectures for the corresponding p -adic congruences. For example, in contrast with (I2 ' ), we conjecture that for any prime p>3 we have the congruences

    \sum\limits_{k = 0}^{p-1} \frac{30k+23}{(-1024)^k} \binom{2k}k \binom{2k}{k+1}T_k(34,1){\equiv} \frac p3 \left(21 \left( \frac 2p \right)-10 \left( \frac{-1}p \right)-11 \right)\ ({\rm{mod}}\ {p^2})

    and

    \sum\limits_{k = 0}^{p-1} \frac{2k+1}{(-1024)^k} \binom{2k}kC_kT_k(34,1){\equiv} \frac p3 \left(2-3 \left( \frac 2p \right)+4 \left( \frac{-1}p \right) \right)\ ({\rm{mod}}\ {p^2}).

    Concerning (I5) and (I5 '' ), we conjecture that

    \frac1{2^{\lfloor n/2\rfloor+1}n \binom{2n}n}\sum\limits_{k = 0}^{n-1}(6k+1) \binom{2k}k^2T_k(8,-2)256^{n-1-k}\in{\Bbb Z}^+

    and

    \frac1{ \binom{2n-2}{n-1}}\sum\limits_{k = 0}^{n-1} \frac{(1-2k-4k^2) \binom{2k}k^2T_k(8,-2)}{(2k-1)^2 256^k}\in{\Bbb Z}^+

    for each n = 2,3,\ldots , and that for any prime p{\equiv}1\ ({\rm{mod}}\ 4) with p = x^2+4y^2\ (x,y\in{\Bbb Z}) we have

    \sum\limits_{k = 0}^{p-1} \frac{ \binom{2k}k^2T_k(8,-2)}{256^k} {\equiv}\left\{ \begin{array}{l} (-1)^{y/2}(4x^2-2p)\ ({\rm{mod}}\ {p^2})& \text{if}\ p{\equiv}1\ ({\rm{mod}}\ 8),\\ (-1)^{(xy-1)/2}8xy\ ({\rm{mod}}\ {p^2})& \text{if}\ p{\equiv}5\ ({\rm{mod}}\ 8), \end{array} \right.

    and

    \sum\limits_{k = 0}^{p-1} \frac{(4k^2+2k-1) \binom{2k}k^2T_k(8,-2)}{(2k-1)^2256^k}{\equiv}0\ ({\rm{mod}}\ {p^2}).

    By [40,Theorem 5.1], we have

    \sum\limits_{k = 0}^{p-1} \frac{ \binom{2k}k^2T_k(8,-2)}{256^k}{\equiv}0\ ({\rm{mod}}\ {p^2})

    for any prime p{\equiv}3\ ({\rm{mod}}\ 4) . The identities (I5), (I5 ' ) and (I5 '' ) were formulated by the author on Dec. 9, 2019.

    Next we pose a conjecture related to the series (Ⅱ1)-(Ⅱ7) and (Ⅱ10)-(Ⅱ12) of Sun [34,40].

    Conjecture 3.2. We have the following identities:

    \sum\limits_{k = 0}^\infty \frac{3k+4}{972^k} \binom{2k}{k+1} \binom{3k}kT_k(18,6) = \frac{63\sqrt3}{40\pi},\;\;\;\;\;\;\;\;(Ⅱ1')
    \sum\limits_{k = 0}^\infty \frac{91k+107}{10^{3k}} \binom{2k}{k+1} \binom{3k}kT_k(10,1) = \frac{275\sqrt3}{18\pi},\;\;\;\;\;\;\;(Ⅱ2')
    \sum\limits_{k = 0}^\infty \frac{195k+83}{18^{3k}} \binom{2k}{k+1} \binom{3k}{k}T_k(198,1) = \frac{9423\sqrt3}{10\pi},\;\;\;\;\;\;\;(Ⅱ3')
    \sum\limits_{k = 0}^\infty \frac{483k-419}{30^{3k}} \binom{2k}{k+1} \binom{3k}kT_k(970,1) = \frac{6550\sqrt3}{\pi},\;\;\;\;\;\;\;(Ⅱ4')
    \sum\limits_{k = 0}^\infty \frac{666k+757}{30^{3k}} \binom{2k}{k+1} \binom{3k}{k}T_k(730,729) = \frac{3475\sqrt3}{4\pi},\;\;\;\;\;\;\;(Ⅱ5')
    \sum\limits_{k = 0}^\infty \frac{8427573k+8442107}{102^{3k}} \binom{2k}{k+1} \binom{3k}kT_k(102,1) = \frac{125137\sqrt6}{20\pi},\;\;\;\;\;\;\;(Ⅱ6')
    \sum\limits_{k = 0}^\infty \frac{959982231k+960422503}{198^{3k}} \binom{2k}{k+1} \binom{3k}kT_k(198,1) = \frac{5335011\sqrt3}{20\pi},\;\;\;\;\;\;\;(Ⅱ7')
    \sum\limits_{k = 0}^\infty \frac{99k+1}{24^{3k}} \binom{2k}{k+1} \binom{3k}kT_k(26,729) = \frac{16(289\sqrt{15}-645\sqrt3)}{15\pi},\;\;\;\;\;\;\;(Ⅱ10')
    \sum\limits_{k = 0}^\infty \frac{45k+1}{(-5400)^k} \binom{2k}{k+1} \binom{3k}kT_k(70,3645) = \frac{345\sqrt3-157\sqrt{15}}{6\pi},\;\;\;\;\;\;\;(Ⅱ11')
    \sum\limits_{k = 0}^\infty \frac{252k-1}{(-13500)^k} \binom{2k}{k+1} \binom{3k}kT_k(40,1458) = \frac{25(1212\sqrt3-859\sqrt6)}{24\pi},\;\;\;\;\;\;\;(Ⅱ12')
    \sum\limits_{k = 0}^\infty \frac{9k+2}{(-675)^k} \binom{2k}{k} \binom{3k}kT_k(15,-5) = \frac{7\sqrt{15}}{8\pi},\;\;\;\;\;\;\;(Ⅱ13)
    \sum\limits_{k = 0}^\infty \frac{45k+31}{(-675)^k} \binom{2k}{k+1} \binom{3k}kT_k(15,-5) = - \frac{19\sqrt{15}}{8\pi},\;\;\;\;\;\;\;(Ⅱ13')
    \sum\limits_{k = 0}^\infty \frac{39k+7}{(-1944)^k} \binom{2k}{k} \binom{3k}kT_k(18,-3) = \frac{9\sqrt{3}}{\pi},\;\;\;\;\;\;\;(Ⅱ14)
    \sum\limits_{k = 0}^\infty \frac{312k+263}{(-1944)^k} \binom{2k}{k+1} \binom{3k}kT_k(18,-3) = - \frac{45\sqrt{3}}{2\pi}.\;\;\;\;\;\;\;(Ⅱ14')

    Remark 3.2. We also have conjectures on related congruences. For example, concerning (Ⅱ), for any prime p>3 we conjecture that

    \sum\limits_{k = 0}^{p-1} \frac{39k+7}{(-1944)^k} \binom{2k}k \binom{3k}kT_k(18,-3){\equiv} \frac p2 \left(13 \left( \frac p3 \right)+1 \right) \ ({\rm{mod}}\ {p^2})

    and that

    \begin{align*} &\sum\limits_{k = 0}^{p-1} \frac{ \binom{2k}k \binom{3k}kT_k(18,-3)}{(-1944)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \text{if}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p7) = 1\ &\ p = x^2+21y^2, \\2p-2x^2\ ({\rm{mod}}\ {p^2})& \text{if}\ ( \frac{-1}p) = ( \frac p3) = -1,\ ( \frac p7) = 1\ &\ 2p = x^2+21y^2, \\12x^2-2p\ ({\rm{mod}}\ {p^2})& \text{if}\ ( \frac{-1}p) = ( \frac p7) = -1,\ ( \frac p3) = 1\ &\ p = 3x^2+7y^2, \\2p-6x^2\ ({\rm{mod}}\ {p^2})& \text{if}\ ( \frac{-1}p) = 1,\ ( \frac p3) = ( \frac p7) = -1\ &\ 2p = 3x^2+7y^2, \\0\ ({\rm{mod}}\ {p^2})& \text{if}\ ( \frac{-21}p) = -1, \end{cases} \end{align*}

    where x and y are integers. The identities (Ⅱ13), (Ⅱ13 ' ), (Ⅱ14) and (Ⅱ14 ' ) were found by the author on Dec. 11, 2019.

    The following conjecture is related to the series (Ⅲ1)-(Ⅲ10) and (Ⅲ12) of Sun [34,40].

    Conjecture 3.3. We have the following identities:

    \sum\limits_{k = 0}^\infty \frac{17k+18}{66^{2k}} \binom{2k}{k+1} \binom{4k}{2k}T_k(52,1) = \frac{77\sqrt{33}}{12\pi},\;\;\;\;\;\;\;\;(Ⅲ1')
    \sum\limits_{k = 0}^\infty \frac{4k+3}{(-96^2)^k} \binom{2k}{k+1} \binom{4k}{2k}T_k(110,1) = - \frac{\sqrt6}{3\pi},\;\;\;\;\;\;\;\;(Ⅲ2')
    \sum\limits_{k = 0}^\infty \frac{8k+9}{112^{2k}} \binom{2k}{k+1} \binom{4k}{2k}T_k(98,1) = \frac{154\sqrt{21}}{135\pi},\;\;\;\;\;\;\;\;(Ⅲ3')
    \sum\limits_{k = 0}^\infty \frac{3568k+4027}{264^{2k}} \binom{2k}{k+1} \binom{4k}{2k}T_k(257,256) = \frac{869\sqrt{66}}{10\pi},\;\;\;\;\;\;\;\;(Ⅲ4')
    \sum\limits_{k = 0}^\infty \frac{144k+1}{(-168^2)^{k}} \binom{2k}{k+1} \binom{4k}{2k}T_k(7,4096) = \frac{7(1745\sqrt{42}-778\sqrt{210})}{120\pi},\;\;\;\;\;\;\;\;(Ⅲ5')
    \sum\limits_{k = 0}^\infty \frac{3496k+3709}{336^{2k}} \binom{2k}{k+1} \binom{4k}{2k}T_k(322,1) = \frac{182\sqrt7}{\pi},\;\;\;\;\;\;\;\;(Ⅲ6')
    \sum\limits_{k = 0}^\infty \frac{286k+229}{336^{2k}} \binom{2k}{k+1} \binom{4k}{2k}T_k(1442,1) = \frac{1113\sqrt{210}}{20\pi},\;\;\;\;\;\;\;\;(Ⅲ7')
    \sum\limits_{k = 0}^\infty \frac{8426k+8633}{912^{2k}} \binom{2k}{k+1} \binom{4k}{2k}T_k(898,1) = \frac{703\sqrt{114}}{20\pi},\;\;\;\;\;\;\;\;(Ⅲ8')
    \sum\limits_{k = 0}^\infty \frac{1608k+79}{912^{2k}} \binom{2k}{k+1} \binom{4k}{2k}T_k(12098,1) = \frac{67849\sqrt{399}}{105\pi},\;\;\;\;\;\;\;\;(Ⅲ9')
    \sum\limits_{k = 0}^\infty \frac{134328722k+134635283}{10416^{2k}} \binom{2k}{k+1} \binom{4k}{2k}T_k(10402,1) = \frac{93961\sqrt{434}}{4\pi},\;\;\;\;\;\;\;\;(Ⅲ10')

    and

    \begin{equation*} \begin{aligned}&\sum\limits_{k = 0}^\infty \frac{39600310408k+39624469807}{39216^{2k}} \binom{2k}{k+1} \binom{4k}{2k}T_k(39202,1) \\&\qquad\qquad = \frac{1334161\sqrt{817}}{\pi}.\end{aligned}\;\;\;\;\;\;\;\;(Ⅲ12') \end{equation*}

    The following conjecture is related to the series (Ⅳ1)-(Ⅳ21) of Sun [34,40].

    Conjecture 3.4. We have the following identities:

    \sum\limits_{k = 0}^\infty \frac{(356k^2+288k+7) \binom{2k}k^2T_{2k}(7,1)}{(k+1)(2k-1)(-48^2)^k} = - \frac{304}{3\pi},\;\;\;\;\;\;\;\;(Ⅳ1')
    \sum\limits_{k = 0}^\infty \frac{(172k^2+141k-1) \binom{2k}k^2T_{2k}(62,1)}{(k+1)(2k-1)(-480^2)^k} = - \frac{80}{3\pi},\;\;\;\;\;\;\;\;(Ⅳ2')
    \sum\limits_{k = 0}^\infty \frac{(782k^2+771k+19) \binom{2k}k^2T_{2k}(322,1)}{(k+1)(2k-1)(-5760^2)^k} = - \frac{90}{\pi},\;\;\;\;\;\;\;\;(Ⅳ3')
    \sum\limits_{k = 0}^\infty \frac{(34k^2+45k+5) \binom{2k}k^2T_{2k}(10,1)}{(k+1)(2k-1)96^{2k}} = - \frac{20\sqrt2}{3\pi},\;\;\;\;\;\;\;\;(Ⅳ4')
    \sum\limits_{k = 0}^\infty \frac{(106k^2+193k+27) \binom{2k}k^2T_{2k}(38,1)}{(k+1)(2k-1)240^{2k}} = - \frac{10\sqrt6}{\pi},\;\;\;\;\;\;\;\;(Ⅳ5')
    \sum\limits_{k = 0}^\infty \frac{(214166k^2+221463k+7227) \binom{2k}k^2T_{2k}(198,1)}{(k+1)(2k-1)39200^{2k}} = - \frac{9240\sqrt6}{\pi},\;\;\;\;\;\;\;\;(Ⅳ6')
    \sum\limits_{k = 0}^\infty \frac{(112k^2+126k+9) \binom{2k}k^2T_{2k}(18,1)}{(k+1)(2k-1)320^{2k}} = - \frac{6\sqrt{15}}{\pi},\;\;\;\;\;\;\;\;(Ⅳ7')
    \sum\limits_{k = 0}^\infty \frac{(926k^2+995k+55) \binom{2k}k^2T_{2k}(30,1)}{(k+1)(2k-1)896^{2k}} = - \frac{60\sqrt7}{\pi},\;\;\;\;\;\;\;\;(Ⅳ8')
    \sum\limits_{k = 0}^\infty \frac{(1136k^2+2962k+503) \binom{2k}k^2T_{2k}(110,1)}{(k+1)(2k-1)24^{4k}} = - \frac{90\sqrt7}{\pi},\;\;\;\;\;\;\;\;(Ⅳ9')
    \sum\limits_{k = 0}^\infty \frac{(5488k^2+8414k+901) \binom{2k}k^2T_{2k}(322,1)}{(k+1)(2k-1)48^{4k}} = - \frac{294\sqrt7}{\pi},\;\;\;\;\;\;\;\;(Ⅳ10')
    \sum\limits_{k = 0}^\infty \frac{(170k^2+193k+11) \binom{2k}k^2T_{2k}(198,1)}{(k+1)(2k-1)2800^{2k}} = - \frac{6\sqrt{14}}{\pi},\;\;\;\;\;\;\;\;(Ⅳ11')
    \sum\limits_{k = 0}^\infty \frac{(104386k^2+108613k+4097) \binom{2k}k^2T_{2k}(102,1)}{(k+1)(2k-1)10400^{2k}} = - \frac{2040\sqrt{39}}{\pi},\;\;\;\;\;\;\;\;(Ⅳ12')
    \sum\limits_{k = 0}^\infty \frac{(7880k^2+8217k+259) \binom{2k}k^2T_{2k}(1298,1)}{(k+1)(2k-1)46800^{2k}} = - \frac{144\sqrt{26}}{\pi},\;\;\;\;\;\;\;\;(Ⅳ13')
    \sum\limits_{k = 0}^\infty \frac{(6152k^2+45391k+9989) \binom{2k}k^2T_{2k}(1298,1)}{(k+1)(2k-1)5616^{2k}} = - \frac{663\sqrt3}{\pi},\;\;\;\;\;\;\;\;(Ⅳ14')
    \sum\limits_{k = 0}^\infty \frac{(147178k^2+2018049k+471431) \binom{2k}k^2T_{2k}(4898,1)}{(k+1)(2k-1)20400^{2k}} = -3740 \frac{\sqrt{51}}{\pi},\;\;\;\;\;\;\;\;(Ⅳ15')
    \sum\limits_{k = 0}^\infty \frac{(1979224k^2+5771627k+991993) \binom{2k}k^2T_{2k}(5778,1)}{(k+1)(2k-1)28880^{2k}} = -73872 \frac{\sqrt{10}}{\pi},\;\;\;\;\;\;\;\;(Ⅳ16')
    \sum\limits_{k = 0}^\infty \frac{(233656k^2+239993k+5827) \binom{2k}k^2T_{2k}(5778,1)}{(k+1)(2k-1)439280^{2k}} = -4080 \frac{\sqrt{19}}{\pi},\;\;\;\;\;\;\;\;(Ⅳ17')
    \sum\limits_{k = 0}^\infty \frac{(5890798k^2+32372979k+6727511) \binom{2k}k^2T_{2k}(54758,1)}{(k+1)(2k-1)243360^{2k}} = -600704 \frac{\sqrt{95}}{9\pi},\;\;\;\;\;\;\;\;(Ⅳ18')
    \sum\limits_{k = 0}^\infty \frac{(148k^2+272k+43) \binom{2k}k^2T_{2k}(10,-2)}{(k+1)(2k-1)4608^{k}} = -28 \frac{\sqrt{6}}{\pi},\;\;\;\;\;\;\;\;(Ⅳ19')
    \sum\limits_{k = 0}^\infty \frac{(3332k^2+17056k+3599) \binom{2k}k^2T_{2k}(238,-14)}{(k+1)(2k-1)1161216^{k}} = -744 \frac{\sqrt{2}}{\pi},\;\;\;\;\;\;\;\;(Ⅳ20')
    \sum\limits_{k = 0}^\infty \frac{(11511872k^2+10794676k+72929) \binom{2k}k^2T_{2k}(9918,-19)}{(k+1)(2k-1)(-16629048064)^{k}} = -390354 \frac{\sqrt{7}}{\pi}.\;\;\;\;\;\;\;\;(Ⅳ21')

    For the five open conjectural series (Ⅵ1), (Ⅵ2), (Ⅵ3), (ⅥI2) and (ⅥI7) of Sun [34,40], we make the following conjecture on related supercongruences.

    Conjecture 3.5. Let p be an odd prime and let n\in{\Bbb Z}^+ . If ( \frac 3p) = 1 , then

    \begin{equation*} \label{VI1} \sum\limits_{k = 0}^{pn-1} \frac{66k+17}{(2^{11}3^3)^k}T_k(10,11^2)^3 -p \left( \frac{-2}p \right)\sum\limits_{k = 0}^{n-1} \frac{66k+17}{(2^{11}3^3)^k}T_k(10,11^2)^3 \end{equation*}

    divided by (pn)^2 is a p -adic integer. If p\not = 5 , then

    \begin{equation*} \label{VI2} \sum\limits_{k = 0}^{pn-1} \frac{126k+31}{(-80)^{3k}}T_k(22,21^2)^3 -p \left( \frac{-5}p \right)\sum\limits_{k = 0}^{n-1} \frac{126k+31}{(-80)^{3k}}T_k(22,21^2)^3 \end{equation*}

    divided by (pn)^2 is a p -adic integer. If ( \frac 7p) = 1 but p\not = 3 , then

    \begin{equation*} \label{VI3} \sum\limits_{k = 0}^{pn-1} \frac{3990k+1147}{(-288)^{3k}}T_k(62,95^2)^3 -p \left( \frac{-2}p \right)\sum\limits_{k = 0}^{n-1} \frac{3990k+1147}{(-288)^{3k}}T_k(62,95^2)^3 \end{equation*}

    divided by (pn)^2 is a p -adic integer. If p{\equiv}\pm1\ ({\rm{mod}}\ 8) but p\not = 7 , then

    \begin{equation*} \label{VII2} \sum\limits_{k = 0}^{pn-1} \frac{24k+5}{28^{2k}} \binom{2k}kT_k(4,9)^2 -p \left( \frac p3 \right)\sum\limits_{k = 0}^{n-1} \frac{24k+5}{28^{2k}} \binom{2k}kT_k(4,9)^2 \end{equation*}

    divided by (pn)^2 is a p -adic integer. If ( \frac{-6}p) = 1 but p\not = 7,31 , then

    \sum\limits_{k = 0}^{pn-1} \frac{2800512k+435257}{434^{2k}} \binom{2k}kT_k(73,576)^2 \\ -p\sum\limits_{k = 0}^{n-1} \frac{2800512k+435257}{434^{2k}} \binom{2k}kT_k(73,576)^2

    divided by (pn)^2 is a p -adic integer.

    Now we pose four conjectural series for 1/\pi of type Ⅷ.

    Conjecture 3.6. We have

    \sum\limits_{k = 0}^\infty \frac{40k+13}{(-50)^k}T_k(4,1)T_k(1,-1)^2 = \frac{55\sqrt{15}}{9\pi},\;\;\;\;\;\;\;\;(Ⅷ1)
    \sum\limits_{k = 0}^\infty \frac{1435k+113}{3240^k}T_k(7,1)T_k(10,10)^2 = \frac{1452\sqrt{5}}{\pi},\;\;\;\;\;\;\;\;(Ⅷ2)
    \sum\limits_{k = 0}^\infty \frac{840k+197}{(-2430)^k}T_k(8,1)T_k(5,-5)^2 = \frac{189\sqrt{15}}{2\pi},\;\;\;\;\;\;\;\;(Ⅷ3)
    \sum\limits_{k = 0}^\infty \frac{39480k+7321}{(-29700)^k}T_k(14,1)T_k(11,-11)^2 = \frac{6795\sqrt{5}}{\pi}.\;\;\;\;\;\;\;\;(Ⅷ4)

    Remark 3.3. The author found the identity (Ⅷ1) on Nov. 3, 2019. The identities (Ⅷ2), (Ⅷ3) and (Ⅷ4) were formulated on Nov. 4, 2019.

    Below we present some conjectures on congruences related to Conjecture 3.6.

    Conjecture 3.7. (ⅰ) For each n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(40k+13)(-1)^k50^{n-1-k}T_k(4,1)T_k(1,-1)^2\in{\Bbb Z}^+, \end{equation} (3.1)

    and this number is odd if and only if n is a power of two ( i.e., n\in\{2^a:\ a\in{\Bbb N}\}) .

    (ⅱ) Let p\not = 2,5 be a prime. Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{40k+13}{(-50)^k}T_k(4,1)T_k(1,-1)^2{\equiv} \frac p3 \left(12+5 \left( \frac3p \right)+22 \left( \frac {-15}p \right) \right) \ ({\rm{mod}}\ {p^2}). \end{equation} (3.2)

    If ( \frac 3p) = ( \frac{-5}p) = 1 , then

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{40k+13}{(-50)^k}T_k(4,1)T_k(1,-1)^2-p\sum\limits_{k = 0}^{n-1} \frac{40k+13}{(-50)^k}T_k(4,1)T_k(1,-1)^2\right) \in{\Bbb Z}_p \end{equation} (3.3)

    for all n\in{\Bbb Z}^+ .

    (ⅲ) Let p\not = 2,5 be a prime. Then

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{T_k(4,1)T_k(1,-1)^2}{(-50)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,9\ ({\rm{mod}}\ {20})\ &\ p = x^2+5y^2\ (x,y\in{\Bbb Z}), \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}3,7\ ({\rm{mod}}\ {20})\ &\ 2p = x^2+5y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-5}p) = -1.\end{cases} \end{aligned} \end{equation} (3.4)

    Remark 3.4. The imaginary quadratic field {\Bbb Q}(\sqrt{-5}) has class number two.

    Conjecture 3.8. (ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(40k+27)(-6)^{n-1-k}T_k(4,1)T_k(1,-1)^2\in{\Bbb Z}, \end{equation} (3.5)

    and the number is odd if and only if n is a power of two.

    (ⅱ) Let p>3 be a prime. Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{40k+27}{(-6)^k}T_k(4,1)T_k(1,-1)^2{\equiv} \frac p9 \left(55 \left( \frac{-5}p \right)+198 \left( \frac 3p \right)-10 \right) \ ({\rm{mod}}\ {p^2}). \end{equation} (3.6)

    If ( \frac 3p) = ( \frac{-5}p) = 1 , then

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{40k+27}{(-6)^k}T_k(4,1)T_k(1,-1)^2-p\sum\limits_{k = 0}^{n-1} \frac{40k+27}{(-6)^k}T_k(4,1)T_k(1,-1)^2\right) \in{\Bbb Z}_p \end{equation} (3.7)

    for all n\in{\Bbb Z}^+ .

    (ⅲ) Let p>5 be a prime. Then

    \begin{equation} \begin{aligned}& \left( \frac p3 \right)\sum\limits_{k = 0}^{p-1} \frac{T_k(4,1)T_k(1,-1)^2}{(-6)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,9\ ({\rm{mod}}\ {20})\ &\ p = x^2+5y^2\ (x,y\in{\Bbb Z}), \\2p-2x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}3,7\ ({\rm{mod}}\ {20})\ &\ 2p = x^2+5y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-5}p) = -1.\end{cases} \end{aligned} \end{equation} (3.8)

    Remark 3.5. This conjecture can be viewed as the dual of Conjecture 3.7. Note that the series \sum_{k = 0}^\infty \frac{(40k+27}{(-6)^k}T_k(4,1)T_k(1,-1)^2 diverges.

    Conjecture 3.9. (ⅰ) For each n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1{n10^{n-1}}\sum\limits_{k = 0}^{n-1}(1435k+113) 3240^{n-1-k}T_k(7,1)T_k(10,10)^2\in{\Bbb Z}^+. \end{equation} (3.9)

    (ⅱ) Let p>3 be a prime. Then

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{1435k+113}{3240^k}T_k(7,1)T_k(10,10)^2 \\{\equiv}& \frac p9 \left(2420 \left( \frac{-5}p \right)+105 \left( \frac5p \right)-1508 \right) \ ({\rm{mod}}\ {p^2}).\end{aligned} \end{equation} (3.10)

    If p{\equiv}1,9\ ({\rm{mod}}\ {20}) , then

    \begin{equation} \sum\limits_{k = 0}^{pn-1} \frac{1435k+113}{3240^k}T_k(7,1)T_k(10,10)^2 -p\sum\limits_{k = 0}^{n-1} \frac{1435k+113}{3240^k}T_k(7,1)T_k(10,10)^2 \end{equation} (3.11)

    divided by (pn)^2 is a p -adic integer for each n\in{\Bbb Z}^+ .

    (ⅲ) Let p>5 be a prime. Then

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{T_k(7,1)T_k(10,10)^2}{3240^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,4\ ({\rm{mod}}\ {15})\ &\ p = x^2+15y^2\ (x,y\in{\Bbb Z}), \\12x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}2,8\ ({\rm{mod}}\ {15})\ &\ p = 3x^2+5y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-15}p) = -1.\end{cases} \end{aligned} \end{equation} (3.12)

    Remark 3.6. The imaginary quadratic field {\Bbb Q}(\sqrt{-15}) has class number two.

    Conjecture 3.10. (ⅰ) For each n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac3{2n10^{n-1}}\sum\limits_{k = 0}^{n-1}(1435k+1322) 50^{n-1-k}T_k(7,1)T_k(10,10)^2\in{\Bbb Z}^+. \end{equation} (3.13)

    (ⅱ) Let p>5 be a prime. Then

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{1435k+1322}{50^k}T_k(7,1)T_k(10,10)^2 \\{\equiv}& \frac p3 \left(3432 \left( \frac{5}p \right)+968 \left( \frac{-1}p \right)-434 \right) \ ({\rm{mod}}\ {p^2}).\end{aligned} \end{equation} (3.14)

    If p{\equiv}1,9\ ({\rm{mod}}\ {20}) , then

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{pn-1} \frac{1435k+1322}{50^k}T_k(7,1)T_k(10,10)^2 -p\sum\limits_{k = 0}^{n-1} \frac{1435k+1322}{50^k}T_k(7,1)T_k(10,10)^2 \end{aligned} \end{equation} (3.15)

    divided by (pn)^2 is a p -adic integer for each n\in{\Bbb Z}^+ .

    (ⅲ) Let p>5 be a prime. Then

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{T_k(7,1)T_k(10,10)^2}{50^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,4\ ({\rm{mod}}\ {15})\ &\ p = x^2+15y^2\ (x,y\in{\Bbb Z}), \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}2,8\ ({\rm{mod}}\ {15})\ &\ p = 3x^2+5y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-15}p) = -1.\end{cases} \end{aligned} \end{equation} (3.16)

    Remark 3.7. This conjecture can be viewed as the dual of Conjecture 3.9. Note that the series

    \sum\limits_{k = 0}^\infty \frac{1435k+1322}{50^k}T_k(7,1)T_k(10,10)^2

    diverges.

    Conjecture 3.11. (ⅰ) For each n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1{n5^{n-1}}\sum\limits_{k = 0}^{n-1}(840k+197)(-1)^k 2430^{n-1-k}T_k(8,1)T_k(5,-5)^2\in{\Bbb Z}^+. \end{equation} (3.17)

    (ⅱ) Let p>3 be a prime. Then

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{840k+197}{(-2430)^k}T_k(8,1)T_k(5,-5)^2 {\equiv} p \left(140 \left( \frac{-15}p \right)+5 \left( \frac{15}p \right)+52 \right) \ ({\rm{mod}}\ {p^2}).\end{aligned} \end{equation} (3.18)

    If ( \frac {-1}p) = ( \frac {15}p) = 1 , then

    \begin{equation} \sum\limits_{k = 0}^{pn-1} \frac{840k+197}{(-2430)^k}T_k(8,1)T_k(5,-5)^2 -p\sum\limits_{k = 0}^{n-1} \frac{840k+197}{(-2430)^k}T_k(8,1)T_k(5,-5)^2 \end{equation} (3.19)

    divided by (pn)^2 is an p -adic integer for any n\in{\Bbb Z}^+ .

    (ⅲ Let p>7 be a prime. Then

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{T_k(8,1)T_k(5,-5)^2}{(-2430)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p5) = ( \frac p7) = 1,\ p = x^2+105y^2, \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p7) = 1,\ ( \frac p3) = ( \frac p5) = -1,\ 2p = x^2+105y^2, \\12x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p5) = ( \frac p7) = -1,\ p = 3x^2+35y^2, \\6x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p7) = -1,\ ( \frac p3) = ( \frac p5) = 1,\ 2p = 3x^2+35y^2, \\2p-20x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p5) = 1,\ ( \frac p5) = ( \frac p7) = -1,\ p = 5x^2+21y^2, \\2p-10x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = 1,\ ( \frac p5) = ( \frac p7) = -1,\ 2p = 5x^2+21y^2, \\28x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p5) = -1,\ ( \frac p3) = ( \frac p7) = 1,\ p = 7x^2+15y^2, \\14x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = -1,\ ( \frac p5) = ( \frac p7) = 1,\ 2p = 7x^2+15y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-105}p) = -1,\end{cases} \end{aligned} \end{equation} (3.20)

    where x and y are integers.

    Remark 3.8. Note that the imaginary quadratic field {\Bbb Q}(\sqrt{-105}) has class number 8 .

    Conjecture 3.12. (ⅰ) For each n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1{n}\sum\limits_{k = 0}^{n-1}(39480k+7321)(-1)^k 29700^{n-1-k}T_k(14,1)T_k(11,-11)^2\in{\Bbb Z}^+, \end{equation} (3.21)

    and this number is odd if and only if n is a power of two.

    (ⅱ) Let p>5 be a prime. Then

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{39480k+7321}{(-29700)^k}T_k(14,1)T_k(11,-11)^2 \\{\equiv}& p \left(5738 \left( \frac{-5}p \right)+70 \left( \frac3p \right)+1513 \right) \ ({\rm{mod}}\ {p^2}).\end{aligned} \end{equation} (3.22)

    If ( \frac 3p) = ( \frac{-5}p) = 1 , then

    \begin{equation} \begin{aligned} &\sum\limits_{k = 0}^{pn-1} \frac{39480k+7321}{(-29700)^k}T_k(14,1)T_k(11,-11)^2 \\&-p\sum\limits_{k = 0}^{n-1} \frac{39480k+7321}{(-29700)^k}T_k(14,1)T_k(11,-11)^2 \end{aligned} \end{equation} (3.23)

    divided by (pn)^2 is a p -adic integer for each n\in{\Bbb Z}^+ .

    (ⅲ) Let p>5 be a prime with p\not = 11 . Then

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{T_k(14,1)T_k(11,-11)^2}{(-29700)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p5) = ( \frac p{11}) = 1,\ p = x^2+165y^2, \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p5) = ( \frac p{11}) = -1,\ 2p = x^2+165y^2, \\12x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p5) = -1,\ ( \frac p3) = ( \frac p{11}) = 1,\ p = 3x^2+55y^2, \\6x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p5) = 1,\ ( \frac p3) = ( \frac p{11}) = -1,\ 2p = 3x^2+55y^2, \\2p-20x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{11}) = 1,\ ( \frac p3) = ( \frac p5) = -1,\ p = 5x^2+33y^2, \\2p-10x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{11}) = -1,\ ( \frac p3) = ( \frac p5) = 1,\ 2p = 5x^2+33y^2, \\44x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = -1,\ ( \frac p5) = ( \frac p{11}) = 1,\ p = 11x^2+15y^2, \\22x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = 1,\ ( \frac p5) = ( \frac p{11}) = -1,\ 2p = 11x^2+15y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-165}p) = -1,\end{cases} \end{aligned} \end{equation} (3.24)

    where x and y are integers.

    Remark 3.9. Note that the imaginary quadratic field {\Bbb Q}(\sqrt{-165}) has class number 8 .

    Conjectures 4.1–4.14 below provide congruences related to (1.88)–(1.97).

    Conjecture 4.1. (ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(7k+3)S_k(1,-6)24^{n-1-k}\in{\Bbb Z}^+. \end{equation} (4.1)

    (ⅱ) Let p>3 be a prime. Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{7k+3}{24^k}S_k(1,-6){\equiv} \frac p2 \left(5 \left( \frac{-2}p \right)+ \left( \frac 6p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (4.2)

    If p{\equiv}1\ ({\rm{mod}}\ 3) , then

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{7k+3}{24^k}S_k(1,-6)-p \left( \frac{-2}p \right)\sum\limits_{k = 0}^{n-1} \frac{7k+3}{24^k}S_k(1,-6)\right) \in{\Bbb Z}_p \end{equation} (4.3)

    for all n\in{\Bbb Z}^+ .

    (ⅲ) For any prime p>3 , we have

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,-6)}{24^k} \\{\equiv}&\begin{cases}( \frac p3)(4x^2-2p)\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,3\ ({\rm{mod}}\ 8)\ &\ p = x^2+2y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}5,7\ ({\rm{mod}}\ 8). \end{cases}\end{aligned} \end{equation} (4.4)

    Conjecture 4.2. (ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(12k+5)S_k(1,7)(-1)^k28^{n-1-k}\in{\Bbb Z}^+, \end{equation} (4.5)

    and this number is odd if and only if n is a power of two.

    (ⅱ) Let p\not = 7 be an odd prime. Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{12k+5}{(-28)^k}S_k(1,7){\equiv}5p \left( \frac p7 \right)\ ({\rm{mod}}\ {p^2}), \end{equation} (4.6)

    and moreover

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{12k+5}{(-28)^k}S_k(1,7)-p \left( \frac p7 \right)\sum\limits_{k = 0}^{n-1} \frac{12k+5}{(-28)^k}S_k(1,7)\right)\in{\Bbb Z}_p \end{equation} (4.7)

    for all n\in{\Bbb Z}^+ .

    (ⅲ) For any prime p\not = 2,7 , we have

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,7)}{(-28)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p7) = 1\ &\ p = x^2+21y^2, \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = -1,\ ( \frac p7) = 1\ &\ 2p = x^2+21y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p7) = -1,\ ( \frac p3) = 1\ &\ p = 3x^2+7y^2, \\2p-6x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = 1,\ ( \frac p3) = ( \frac p7) = -1\ &\ 2p = 3x^2+7y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-21}p) = -1, \end{cases}\end{aligned} \end{equation} (4.8)

    where x and y are integers.

    Conjecture 4.3. (ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(84k+29)S_k(1,-20)80^{n-1-k}\in{\Bbb Z}^+, \end{equation} (4.9)

    and this number is odd if and only if n is a power of two.

    (ⅱ) Let p be an odd prime with p\not = 5 . Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{84k+29}{80^k}S_k(1,-20){\equiv} p \left(2 \left( \frac 5p \right)+27 \left( \frac{-15}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (4.10)

    If p{\equiv}1\ ({\rm{mod}}\ 3) , then

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{84k+29}{80^k}S_k(1,-20)- p \left( \frac p5 \right)\sum\limits_{k = 0}^{n-1} \frac{84k+29}{80^k}S_k(1,-20)\right)\in{\Bbb Z}_p \end{equation} (4.11)

    for all n\in{\Bbb Z}^+ .

    (ⅲ) For any prime p\not = 2,5 , we have

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,-20)}{80^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p3) = ( \frac p5) = 1\ &\ p = x^2+30y^2, \\8x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = 1,\ ( \frac p3) = ( \frac p5) = -1\ &\ p = 2x^2+15y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p3) = 1,\ ( \frac 2p) = ( \frac p5) = -1\ &\ p = 3x^2+10y^2, \\20x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{p}5) = 1,\ ( \frac 2p) = ( \frac p3) = -1\ &\ p = 5x^2+6y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-30}p) = -1, \end{cases}\end{aligned} \end{equation} (4.12)

    where x and y are integers.

    Conjecture 4.4. (ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(3k+1)(-1)^k100^{n-1-k}S_k(1,25)\in{\Bbb Z}^+. \end{equation} (4.13)

    (ⅱ) Let p\not = 5 be an odd prime. Then

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{3k+1}{(-100)^k}S_k(1,25)- p \left( \frac{-1}p \right)\sum\limits_{k = 0}^{n-1} \frac{3k+1}{(-100)^k}S_k(1,25)\right)\in{\Bbb Z}_p \end{equation} (4.14)

    for all n\in{\Bbb Z}^+ .

    (ⅲ) For any prime p>3 with p\not = 11 , we have

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,25)}{(-100)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p{11}) = 1\ &\ p = x^2+33y^2, \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}{p}) = 1,\ ( \frac p3) = ( \frac p{11}) = -1\ &\ 2p = x^2+33y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p{11}) = 1,\ ( \frac {-1}p) = ( \frac p3) = -1\ &\ p = 3x^2+11y^2, \\2p-6x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{p}3) = 1,\ ( \frac {-1}p) = ( \frac p{11}) = -1\ &\ 2p = 3x^2+11y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-33}p) = -1, \end{cases}\end{aligned} \end{equation} (4.15)

    where x and y are integers.

    Conjecture 4.5. (ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(228k+67)S_k(1,-56)224^{n-1-k}\in{\Bbb Z}^+, \end{equation} (4.16)

    and this number is odd if and only if n is a power of two.

    (ⅱ) Let p be an odd prime with p\not = 7 . Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{228k+67}{224^k}S_k(1,-56){\equiv} p \left(65 \left( \frac{-7}p \right)+2 \left( \frac {14}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (4.17)

    If p{\equiv}1,3\ ({\rm{mod}}\ 8) , then

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{228k+67}{224^k}S_k(1,-56)- p \left( \frac p7 \right)\sum\limits_{k = 0}^{n-1} \frac{228k+67}{224^k}S_k(1,-56)\right)\in{\Bbb Z}_p \end{equation} (4.18)

    for all n\in{\Bbb Z}^+ .

    (ⅲ) For any prime p\not = 2,7 , we have

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,-56)}{224^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-2}p) = ( \frac p3) = ( \frac p7) = 1\ &\ p = x^2+42y^2, \\8x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{p}7) = 1,\ ( \frac {-2}p) = ( \frac p3) = -1\ &\ p = 2x^2+21y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac {-2}p) = 1,\ ( \frac p3) = ( \frac p7) = -1\ &\ p = 3x^2+14y^2, \\2p-24x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{p}5) = 1,\ ( \frac 2p) = ( \frac p3) = -1\ &\ p = 6x^2+7y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-42}p) = -1, \end{cases}\end{aligned} \end{equation} (4.19)

    where x and y are integers.

    Conjecture 4.6. (ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(399k+101)(-1)^k676^{n-1-k}S_k(1,169)\in{\Bbb Z}^+. \end{equation} (4.20)

    (ⅱ) Let p\not = 13 be an odd prime. Then

    \begin{equation} \sum\limits_{k = 0}^{pn-1} \frac{399k+101}{(-676)^k}S_k(1,169)- p \left( \frac{-1}p \right)\sum\limits_{k = 0}^{n-1} \frac{399k+101}{(-676)^k}S_k(1,169) \end{equation} (4.21)

    divided by (pn)^2 is a p -adic integer for any n\in{\Bbb Z}^+ .

    (ⅲ) For any prime p>3 with p\not = 19 , we have

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,169)}{(-676)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p{19}) = 1\ &\ p = x^2+57y^2, \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}{p}) = 1,\ ( \frac p3) = ( \frac p{19}) = -1\ &\ 2p = x^2+57y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p{3}) = 1,\ ( \frac {-1}p) = ( \frac p{19}) = -1\ &\ p = 3x^2+19y^2, \\2p-6x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{p}{19}) = 1,\ ( \frac {-1}p) = ( \frac p{3}) = -1\ &\ 2p = 3x^2+19y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-57}p) = -1, \end{cases}\end{aligned} \end{equation} (4.22)

    where x and y are integers.

    Conjecture 4.7. (ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(2604k+563)S_k(1,-650)2600^{n-1-k}\in{\Bbb Z}^+, \end{equation} (4.23)

    and this number is odd if and only if n\in\{2^a:\ a\in{\Bbb N}\} .

    (ⅱ) Let p be an odd prime with p\not = 5,13 . Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{2604k+563}{2600^k}S_k(1,-650){\equiv} p \left(561 \left( \frac{-39}p \right)+2 \left( \frac {26}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (4.24)

    If ( \frac{-6}p) = 1 , then

    \begin{equation} \sum\limits_{k = 0}^{pn-1} \frac{2604k+563}{2600^k}S_k(1,-650)- p \left( \frac{26}p \right)\sum\limits_{k = 0}^{n-1} \frac{2604k+563}{2600^k}S_k(1,-650) \end{equation} (4.25)

    divided by (pn)^2 is a p -adic integer for any n\in{\Bbb Z}^+ .

    (ⅲ)For any odd prime p\not = 5,13 , we have

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,-650)}{2600^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p3) = ( \frac p{13}) = 1\ &\ p = x^2+78y^2, \\8x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac2{p}) = 1,\ ( \frac p3) = ( \frac p{13}) = -1\ &\ p = 2x^2+39y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p{13}) = 1,\ ( \frac 2p) = ( \frac p3) = -1\ &\ p = 3x^2+26y^2, \\2p-24x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{p}3) = 1,\ ( \frac 2p) = ( \frac p{13}) = -1\ &\ p = 6x^2+13y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-78}p) = -1, \end{cases}\end{aligned} \end{equation} (4.26)

    where x and y are integers.

    Conjecture 4.8. (ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(39468k+7817)(-1)^k6076^{n-1-k}S_k(1,1519)\in{\Bbb Z}^+, \end{equation} (4.27)

    and this number is odd if and only if n\in\{2^a:\ a\in{\Bbb N}\} .

    (ⅱ) Let p\not = 7,31 be an odd prime. Then

    \begin{equation} \sum\limits_{k = 0}^{pn-1} \frac{39468k+7817}{(-6076)^k}S_k(1,1519)- p \left( \frac{-31}p \right)\sum\limits_{k = 0}^{n-1} \frac{39468k+7817}{(-6076)^k}S_k(1,1519) \end{equation} (4.28)

    divided by (pn)^2 is a p -adic integer for any n\in{\Bbb Z}^+ .

    (ⅲ) For any prime p>3 with p\not = 7,31 , we have

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,1519)}{(-6076)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p{31}) = 1\ &\ p = x^2+93y^2, \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{p}{31}) = 1,\ ( \frac {-1}p) = ( \frac p{3}) = -1\ &\ 2p = x^2+93y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p{3}) = 1,\ ( \frac {-1}p) = ( \frac p{31}) = -1\ &\ p = 3x^2+31y^2, \\2p-6x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = 1,\ ( \frac {p}3) = ( \frac p{31}) = -1\ &\ 2p = 3x^2+31y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-93}p) = -1, \end{cases}\end{aligned} \end{equation} (4.29)

    where x and y are integers.

    Conjecture 4.9. (ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(41667k+7879)9800^{n-1-k}S_k(1,-2450)\in{\Bbb Z}^+. \end{equation} (4.30)

    (ⅱ) Let p\not = 5,7 be an odd prime. Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{41667k+7879}{9800^k}S_k(1,-2450) {\equiv} \frac p2 \left(15741 \left( \frac{-6}p \right)+17 \left( \frac 2p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (4.31)

    If p{\equiv}1\ ({\rm{mod}}\ 3) , then

    \begin{equation} \sum\limits_{k = 0}^{pn-1} \frac{41667k+7879}{9800^k}S_k(1,-2450)- p \left( \frac{2}p \right)\sum\limits_{k = 0}^{n-1} \frac{41667k+7879}{9800^k}S_k(1,-2450) \end{equation} (4.32)

    divided by (pn)^2 is a p -adic integer for any n\in{\Bbb Z}^+ .

    (ⅲ) For any prime p>7 with p\not = 17 , we have

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,-2450)}{9800^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p3) = ( \frac p{17}) = 1\ &\ p = x^2+102y^2, \\8x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{p}{17}) = 1,\ ( \frac {2}p) = ( \frac p{3}) = -1\ &\ p = 2x^2+51y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p{3}) = 1,\ ( \frac {2}p) = ( \frac p{17}) = -1\ &\ p = 3x^2+34y^2, \\2p-24x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = 1,\ ( \frac {p}3) = ( \frac p{17}) = -1\ &\ p = 6x^2+17y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-102}p) = -1, \end{cases}\end{aligned} \end{equation} (4.33)

    where x and y are integers.

    Conjecture 4.10. (ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(74613k+10711)(-1)^k530^{2(n-1-k)}S_k(1,265^2)\in{\Bbb Z}^+. \end{equation} (4.34)

    \rm(ⅱ) Let p\not = 5,53 be an odd prime. Then

    \begin{equation} \sum\limits_{k = 0}^{pn-1} \frac{74613k+10711}{(-530^2)^k}S_k(1,265^2)- p \left( \frac{-1}p \right)\sum\limits_{k = 0}^{n-1} \frac{74613k+10711}{(-530^2)^k}S_k(1,265^2) \end{equation} (4.35)

    divided by (pn)^2 is a p -adic integer for any n\in{\Bbb Z}^+ .

    \rm(ⅲ) For any prime p>5 with p\not = 59 , we have

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,265^2)}{(-530^2)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p{59}) = 1\ &\ p = x^2+177y^2, \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}{p}) = 1,\ ( \frac {p}3) = ( \frac p{59}) = -1\ &\ 2p = x^2+177y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p{59}) = 1,\ ( \frac {-1}p) = ( \frac p{3}) = -1\ &\ p = 3x^2+59y^2, \\2p-6x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{p}3) = 1,\ ( \frac {-1}p) = ( \frac p{59}) = -1\ &\ 2p = 3x^2+59y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-177}p) = -1, \end{cases}\end{aligned} \end{equation} (4.36)

    where x and y are integers.

    Conjecture 4.11. For any odd prime p ,

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k}{(-4)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ 12\mid p-1\ &\ p = x^2+y^2\ (x,y\in{\Bbb Z}\ &\ 3\nmid x), \\4xy\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ 12\mid p-5\ &\ p = x^2+y^2\ (x,y\in{\Bbb Z}\ &\ 3\mid x-y), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}3\ ({\rm{mod}}\ 4). \end{cases}\end{aligned} \end{equation} (4.37)

    Also, for any prime p{\equiv}1\ ({\rm{mod}}\ 4) we have

    \begin{equation} \sum\limits_{k = 0}^{p-1}(8k+5) \frac{S_k}{(-4)^k}{\equiv}4p\ ({\rm{mod}}\ {p^2}). \end{equation} (4.38)

    Conjecture 4.12. \rm(ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(4k+3)4^{n-1-k}S_k(1,-1)\in{\Bbb Z}, \end{equation} (4.39)

    and this number is odd if and only if n is a power of two.

    \rm(ⅱ) For any odd prime p and positive integer n , we have

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{4k+3}{4^k}S_k(1,-1)- p\sum\limits_{k = 0}^{n-1} \frac{4k+3}{4^k}S_k(1,-1)\right)\in{\Bbb Z}_p. \end{equation} (4.40)

    \rm(ⅲ) Let p be an odd prime. Then

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,-1)}{4^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,9\ ({\rm{mod}}\ {20})\ &\ p = x^2+5y^2\ (x,y\in{\Bbb Z}), \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}3,7\ ({\rm{mod}}\ {20})\ &\ 2p = x^2+5y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-5}p) = -1. \end{cases}\end{aligned} \end{equation} (4.41)

    Conjecture 4.13. \rm(ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(33k+25)S_k(1,-6)(-6)^{n-1-k}\in{\Bbb Z}, \end{equation} (4.42)

    and this number is odd if and only if n is a power of two.

    \rm(ⅱ) Let p>3 be a prime. Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{33k+25}{(-6)^k}S_k(1,-6){\equiv} p \left(35-10 \left( \frac 3p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (4.43)

    If p{\equiv}\pm1\ ({\rm{mod}}\ {12}) , then

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{33k+25}{(-6)^k}S_k(1,-6) -p\sum\limits_{k = 0}^{n-1} \frac{33k+25}{(-6)^k}S_k(1,-6)\right)\in{\Bbb Z}_p \end{equation} (4.44)

    for all n\in{\Bbb Z}^+ .

    \rm(ⅲ) For any prime p>3 , we have

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{S_k(1,-6)}{(-6)^k}{\equiv}\begin{cases}( \frac{-1}p)(4x^2-2p)\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p = x^2+3y^2\ (x,y\in{\Bbb Z}),\\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}2\ ({\rm{mod}}\ 3).\end{cases} \end{equation} (4.45)

    Conjecture 4.14. \rm(ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} n\ \ \bigg| \ \sum\limits_{k = 0}^{n-1}(18k+13)S_k(2,9)8^{n-1-k}. \end{equation} (4.46)

    \rm(ⅱ) Let p be an odd prime. Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{18k+13}{8^k}S_k(2,9) {\equiv} p \left(1+12 \left( \frac p3 \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (4.47)

    If p{\equiv}1\ ({\rm{mod}}\ 3) , then

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{18k+13}{8^k}S_k(2,9)-p\sum\limits_{k = 0}^{n-1} \frac{18k+13}{8^k}S_k(2,9)\right)\in{\Bbb Z}_p \end{equation} (4.48)

    for all n\in{\Bbb Z}^+ .

    \rm(ⅲ) For any prime p>3 , we have

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,-2)}{8^k}{\equiv} \left( \frac p3 \right)\sum\limits_{k = 0}^{p-1} \frac{S_k(2,9)}{8^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,7\ ({\rm{mod}}\ {24})\ &\ p = x^2+6y^2\ \\8x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}5,11\ ({\rm{mod}}\ {24})\ &\ p = 2x^2+3y^2,\\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-6}p) = -1,\end{cases} \end{aligned} \end{equation} (4.49)

    where x and y are integers.

    Conjecture 4.15. Let p be an odd prime with p\not = 5 . Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{S_k(3,1)}{4^k} {\equiv}\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,9\ ({\rm{mod}}\ {20})\ &\ p = x^2+5y^2,\ \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}3,7\ ({\rm{mod}}\ {20})\ &\ 2p = x^2+5y^2,\\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}11,13,17,19\ ({\rm{mod}}\ {20}),\end{cases} \end{equation} (4.50)

    where x and y are integers. If ( \frac{-5}p) = 1 , then

    \sum\limits_{k = 0}^{p-1} \frac{40k+29}{4^k}S_k(3,1){\equiv} 18p\ ({\rm{mod}}\ {p^2}).

    Remark 4.1. We also have some similar conjectures involving

    \begin{gather*} \sum\limits_{k = 0}^{p-1} \frac{S_k(5,4)}{4^k},\ \sum\limits_{k = 0}^{p-1} \frac{S_k(4,-5)}{4^k}, \ \sum\limits_{k = 0}^{p-1} \frac{S_k(7,6)}{6^k}, \\ \sum\limits_{k = 0}^{p-1} \frac{S_k(10,-2)}{32^k}, \ \sum\limits_{k = 0}^{p-1} \frac{S_k(14,9)}{72^k},\ \sum\limits_{k = 0}^{p-1} \frac{S_k(19,9)}{36^k} \end{gather*}

    modulo p^2 , where p is a prime greater than 3 .

    Motivated by Theorem 2.6, we pose the following general conjecture.

    Conjecture 4.16. For any odd prime p and integer m\not{\equiv}0\ ({\rm{mod}}\ p) , we have

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{S_k(4,-m)}{m^k}{\equiv}\sum\limits_{k = 0}^{p-1} \frac{ \binom{2k}kf_k}{m^k}\ ({\rm{mod}}\ {p^2}). \end{equation} (4.51)

    and

    \begin{equation} \frac{m+16}2\sum\limits_{k = 0}^{p-1} \frac{kS_k(4,-m)}{m^k} -\sum\limits_{k = 0}^{p-1}((m+4)k-4) \frac{ \binom{2k}kf_k}{m^k}{\equiv}4p \left( \frac mp \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (4.52)

    Remark 4.2 We have checked this conjecture via \mathsf{Mathematica}. In view of the proof of Theorem 2.6, both (4.51) and (4.52) hold modulo p .

    The numbers

    Z_n: = \sum\limits_{k = 0}^n \binom nk \binom{2k}k \binom{2(n-k)}{n-k}\ \ (n = 0,1,2,\ldots)

    were first introduced by D. Zagier in his paper [51] the preprint of which was released in 2002. Thus we name such numbers as Zagier numbers. As pointed out by the author [41,Remark 4.3], for any n\in{\Bbb N} the number 2^nZ_n coincides with the so-called CLF (Catalan-Larcombe-French) number

    {\mathcal P}_n: = 2^n\sum\limits_{k = 1}^{\lfloor n/2\rfloor} \binom n{2k} \binom{2k}k^24^{n-2k} = \sum\limits_{k = 0}^n \frac{ \binom{2k}k^2 \binom{2(n-k)}{n-k}^2}{ \binom nk}.

    Let p be an odd prime. For any k = 0,\ldots,p-1 , we have

    {\mathcal P}_k{\equiv} \left( \frac{-1}p \right)128^k{\mathcal P}_{p-1-k}\ ({\rm{mod}}\ p)

    by F. Jarvis and H.A. Verrill [24,Corollary 2.2], and hence

    Z_k = \frac{{\mathcal P}_k}{2^k}{\equiv} \left( \frac{-1}p \right)64^k(2^{p-1-k}Z_{p-1-k}){\equiv} \left( \frac{-1}p \right)32^kZ_{p-1-k}\ ({\rm{mod}}\ p).

    Combining this with Remark 1.3(ⅱ), we see that

    \begin{align*} \sum\limits_{k = 0}^{p-1} \frac{Z_kT_k(b,c)}{m^k}{\equiv}& \left( \frac{4c-b^2}p \right)\sum\limits_{k = 0}^{p-1} \left( \frac{32(b^2-4c)}m \right)^kZ_{p-1-k}T_{p-1-k}(b,c) \\{\equiv}& \left( \frac{4c-b^2}p \right)\sum\limits_{k = 0}^{p-1} \frac{Z_kT_k(b,c)}{(32(b^2-4c)/m)^k}\ ({\rm{mod}}\ p) \end{align*}

    for any b,c,m\in{\Bbb Z} with p\nmid (b^2-4c)m .

    J. Wan and Zudilin [49] obtained the following irrational series for 1/\pi involving the Legendre polynomials and the Zagier numbers:

    \sum\limits_{k = 0}^\infty(15k+4-2\sqrt6)Z_kP_k \left( \frac{24-\sqrt6}{15\sqrt2} \right) \left( \frac{4-\sqrt6}{10\sqrt3} \right)^k = \frac{6}{\pi}(7+3\sqrt6).

    Via our congruence approach (including Conjecture 1.4), we find 24 rational series for 1/\pi involving T_n(b,c) and the Zagier numbers. Theorem 1 of [49] might be helpful to solve some of them.

    Conjecture 5.1. We have the following identities for 1/\pi .

    \begin{align} \sum\limits_{k = 1}^\infty \frac{5k+1}{32^k}T_kZ_k& = \frac{8(2+\sqrt5)}{3\pi}, \end{align} (5.1)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{21k+5}{(-252)^k}T_k(1,16)Z_k& = \frac{6\sqrt7}{\pi}, \end{align} (5.2)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{3k+1}{36^k}T_k(1,-2)Z_k& = \frac{3}{\pi}, \end{align} (5.3)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{k}{192^k}T_k(14,1)Z_k& = \frac{8}{3\pi}, \end{align} (5.4)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{30k+11}{(-192)^k}T_k(14,1)Z_k& = \frac{12}{\pi}, \end{align} (5.5)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{15k+1}{480^k}T_k(22,1)Z_k& = \frac{6\sqrt{10}}{\pi}, \end{align} (5.6)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{7k+2}{(-672)^k}T_k(26,1)Z_k& = \frac{2\sqrt{21}}{3\pi}, \end{align} (5.7)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{21k+2}{1152^k}T_k(34,1)Z_k& = \frac{18}{\pi}, \end{align} (5.8)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{30k-7}{640^k}T_k(62,1)Z_k& = \frac{160}{\pi}, \end{align} (5.9)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{195k+34}{(-9600)^k}T_k(98,1)Z_k& = \frac{80}{\pi}, \end{align} (5.10)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{195k+22}{11232^k}T_k(106,1)Z_k& = \frac{27\sqrt{13}}{\pi}, \end{align} (5.11)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{42k+17}{(-1440)^k}T_k(142,1)Z_k& = \frac{33}{\sqrt5\,\pi}, \end{align} (5.12)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{2k-1}{1792^k}T_k(194,1)Z_k& = \frac{56}{3\pi}, \end{align} (5.13)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{1785k+254}{(-37632)^k}T_k(194,1)Z_k& = \frac{672}{\pi}, \end{align} (5.14)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{210k+23}{40800^k}T_k(202,1)Z_k& = \frac{15\sqrt{34}}{\pi}, \end{align} (5.15)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{210k-1}{4608^k}T_k(254,1)Z_k& = \frac{288}{\pi}, \end{align} (5.16)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{21k-5}{5600^k}T_k(502,1)Z_k& = \frac{105}{\sqrt2\,\pi}, \end{align} (5.17)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{7410k+1849}{(-36992)^k}T_k(1154,1)Z_k& = \frac{2992}{\pi}, \end{align} (5.18)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{1326k+101}{57760^k}T_k(1442,1)Z_k& = \frac{2014}{\sqrt5\,\pi}, \end{align} (5.19)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{78k-131}{20800^k}T_k(2498,1)Z_k& = \frac{2600}{\pi}, \end{align} (5.20)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{62985k+11363}{(-394272)^k}T_k(5474,1)Z_k& = \frac{7659\sqrt{10}}{\pi}, \end{align} (5.21)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{358530k+33883}{486720^k}T_k(6082,1)Z_k& = \frac{176280}{\pi}, \end{align} (5.22)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{510k-1523}{78400^k}T_k(9602,1)Z_k& = \frac{33320}{\pi}, \end{align} (5.23)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{570k-457}{93600^k}T_k(10402,1)Z_k& = \frac{1590\sqrt{13}}{\pi}. \end{align} (5.24)

    Below we present some conjectures on congruences related to (5.1) , (5.2) , (5.4) and (5.9) .

    Conjecture 5.2. (ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} n\ \bigg|\ \sum\limits_{k = 0}^{n-1}(5k+1)T_kZ_k32^{n-1-k}. \end{equation} (5.25)

    \rm (ⅱ) Let p be an odd prime with p\not = 5 . Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{5k+1}{32^k}T_kZ_k{\equiv} \frac p3 \left(5 \left( \frac{-5}p \right)-2 \left( \frac{-1}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (5.26)

    If p{\equiv}\pm1\ ({\rm{mod}}\ {5}) , then

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{5k+1}{32^k}T_kZ_k-p \left( \frac{-1}p \right)\sum\limits_{k = 0}^{n-1} \frac{5k+1}{32^k}T_kZ_k\right) \in{\Bbb Z}_p \end{equation} (5.27)

    for all n\in{\Bbb Z}^+ .

    \rm (ⅲ) For any prime p>5 , we have

    \begin{equation} \begin{aligned}& \left( \frac{-1}p \right)\sum\limits_{k = 0}^{p-1} \frac{T_kZ_k}{32^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,4\ ({\rm{mod}}\ {15})\ &\ p = x^2+15y^2\ (x,y\in{\Bbb Z}), \\12x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}2,8\ ({\rm{mod}}\ {15})\ &\ p = 3x^2+5y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-15}p) = -1. \end{cases} \end{aligned} \end{equation} (5.28)

    Conjecture 5.3. (ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(-1)^k(21k+5)T_k(1,16)Z_k252^{n-1-k}\in{\Bbb Z}^+. \end{equation} (5.29)

    \rm (ⅱ) Let p>3 be a prime with p\not = 7 . Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{21k+5}{(-252)^k}T_k(1,16)Z_k{\equiv} \frac p3 \left(16 \left( \frac{-7}p \right)- \left( \frac{-1}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (5.30)

    If ( \frac 7p) = 1 , then

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{21k+5}{(-252)^k}T_k(1,16)Z_k-p \left( \frac{-1}p \right) \sum\limits_{k = 0}^{n-1} \frac{21k+5}{(-252)^k}T_k(1,16)Z_k\right)\in{\Bbb Z}_p \end{equation} (5.31)

    for all n\in{\Bbb Z}^+ .

    \rm (ⅲ) For any prime p>3 with p\not = 7 , we have

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{T_k(1,16)Z_k}{(-252)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,2,4\ ({\rm{mod}}\ {7})\ &\ p = x^2+7y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv} 3,5,6\ ({\rm{mod}}\ 7). \end{cases} \end{aligned} \end{equation} (5.32)

    Conjecture 5.4. \rm(i) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} n\ \bigg|\ \sum\limits_{k = 0}^{n-1}kT_k(14,1)Z_k192^{n-1-k}. \end{equation} (5.33)

    \rm (ⅱ) Let p>3 be a prime. Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{k}{192^k}T_k(14,1)Z_k{\equiv} \frac p9 \left( \left( \frac{-1}p \right)- \left( \frac{2}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (5.34)

    If p{\equiv}1,3\ ({\rm{mod}}\ 8) , then

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{k}{192^k}T_k(14,1)Z_k-p \left( \frac{-1}p \right) \sum\limits_{k = 0}^{n-1} \frac{k}{192^k}T_k(14,1)\right)\in{\Bbb Z}_p \end{equation} (5.35)

    for all n\in{\Bbb Z}^+ .

    \rm (ⅲ) For any prime p>3 , we have

    \begin{equation} \begin{aligned}& \left( \frac 3p \right)\sum\limits_{k = 0}^{p-1} \frac{T_k(14,1)Z_k}{192^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,3\ ({\rm{mod}}\ {8})\ &\ p = x^2+2y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv} 5,7\ ({\rm{mod}}\ 8). \end{cases} \end{aligned} \end{equation} (5.36)

    Conjecture 5.5. \rm(ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} n\ \bigg|\ \sum\limits_{k = 0}^{n-1}(30k-7)T_k(62,1)Z_k640^{n-1-k}. \end{equation} (5.37)

    \rm (ⅱ) Let p be an odd prime with p\not = 5 . Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{30k-7}{640^k}T_k(62,1)Z_k{\equiv} p \left(2 \left( \frac{-1}p \right)-9 \left( \frac{15}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (5.38)

    If ( \frac{-15}p) = 1 , then

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{30k-7}{640^k}T_k(62,1)Z_k -p \left( \frac{-1}p \right)\sum\limits_{k = 0}^{n-1} \frac{30k-7}{640^k}T_k(62,1)Z_k\right)\in{\Bbb Z}_p \end{equation} (5.39)

    for all n\in{\Bbb Z}^+ .

    \rm (ⅲ) For any prime p>5 , we have

    \begin{equation} \begin{aligned}& \left( \frac{-1}p \right)\sum\limits_{k = 0}^{p-1} \frac{T_k(62,1)Z_k}{640^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac 2p) = ( \frac p3) = ( \frac p5) = 1\ &\ p = x^2+30y^2, \\8x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac 2p) = 1,\ ( \frac p3) = ( \frac p5) = -1\ &\ p = 2x^2+15y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p3) = 1,\ ( \frac 2p) = ( \frac p5) = -1\ &\ p = 3x^2+10y^2, \\20x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p5) = 1,\ ( \frac 2p) = ( \frac p3) = -1\ &\ p = 5x^2+6y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-30}p) = -1, \end{cases} \end{aligned} \end{equation} (5.40)

    where x and y are integers.

    Sun [36,37] obtained some supercongruences involving the Franel numbers f_n = \sum_{k = 0}^n \binom nk^3\ (n\in{\Bbb N}) . M. Rogers and A. Straub [30] confirmed the 520 -series for 1/\pi involving Franel polynomials conjectured by Sun [34].

    Let p be an odd prime. By [24,Lemma 2.6], we have f_k{\equiv}(-8)^kf_{p-1-k}\ ({\rm{mod}}\ p) for each k = 0,\ldots,p-1 . Combining this with Remark 1.3(ⅱ), we see that

    \begin{align*} \sum\limits_{k = 0}^{p-1} \frac{f_kT_k(b,c)}{m^k}{\equiv}& \left( \frac{b^2-4c}p \right)\sum\limits_{k = 0}^{p-1} \left( \frac{-8(b^2-4c)}m \right)^kf_{p-1-k}T_{p-1-k}(b,c) \\{\equiv}& \left( \frac{b^2-4c}p \right)\sum\limits_{k = 0}^{p-1} \frac{f_kT_k(b,c)}{(8(4c-b^2)/m)^k}\ ({\rm{mod}}\ p) \end{align*}

    for any b,c,m\in{\Bbb Z} with p\nmid (b^2-4c)m .

    Wan and Zudilin [49] deduced the following irrational series for 1/\pi involving the Legendre polynomials and the Franel numbers:

    \sum\limits_{k = 0}^\infty(18k+7-2\sqrt3)f_kP_k \left( \frac{1+\sqrt3}{\sqrt6} \right) \left( \frac{2-\sqrt3}{2\sqrt6} \right)^k = \frac{27+11\sqrt3}{\sqrt2\,\pi}.

    Via our congruence approach (including Conjecture 1.4), we find 12 rational series for 1/\pi involving T_n(b,c) and the Franel numbers; Theorem 1 of [49] might be helpful to solve some of them.

    Conjecture 6.1. We have

    \begin{align} \sum\limits_{k = 0}^\infty \frac{3k+1}{(-48)^k}f_kT_k(4,-2)& = \frac{4\sqrt2}{3\pi}, \end{align} (6.1)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{99k+23}{(-288)^k}f_kT_k(8,-2)& = \frac{39\sqrt2}{\pi}, \end{align} (6.2)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{105k+17}{480^k}f_kT_k(8,1)& = \frac{92\sqrt5}{3\pi}, \end{align} (6.3)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{45k-2}{441^k}f_kT_k(47,1)& = \frac{483\sqrt5}{4\pi}, \end{align} (6.4)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{165k+46}{(-2352)^k}f_kT_k(194,1)& = \frac{112\sqrt5}{3\pi}, \end{align} (6.5)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{42k+5}{11616^k}f_kT_k(482,1)& = \frac{374\sqrt2}{15\pi}, \end{align} (6.6)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{990k+31}{11200^k}f_kT_k(898,1)& = \frac{680\sqrt7}{\pi}, \end{align} (6.7)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{585k+172}{(-13552)^k}f_kT_k(1454,1)& = \frac{110\sqrt7}{\pi}, \end{align} (6.8)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{90k+11}{101568^k}f_kT_k(2114,1)& = \frac{92\sqrt{15}}{7\pi}, \end{align} (6.9)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{94185k+17014}{(-105984)^k}f_kT_k(2302,1)& = \frac{8520\sqrt{23}}{\pi}, \end{align} (6.10)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{5355k+1381}{(-61952)^k}f_kT_k(4354,1)& = \frac{968\sqrt{7}}{\pi}, \end{align} (6.11)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{210k+23}{475904^k}f_kT_k(16898,1)& = \frac{2912\sqrt{231}}{297\pi}. \end{align} (6.12)

    We now present a conjecture on congruence related to (6.3) .

    Conjecture 6.2. \rm(ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(105k+17)480^{n-1-k}f_kT_k(8,1)\in{\Bbb Z}^+. \end{equation} (6.13)

    \rm (ⅱ) Let p>5 be a prime. Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{105k+17}{480^k}f_kT_k(8,1) {\equiv} \frac p9 \left(161 \left( \frac{-5}p \right)-8 \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (6.14)

    If ( \frac{-5}p) = 1 , then

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{105k+17}{480^k}f_kT_k(8,1) -p\sum\limits_{k = 0}^{n-1} \frac{105k+17}{480^k}f_kT_k(8,1)\right)\in{\Bbb Z}_p \end{equation} (6.15)

    for all n\in{\Bbb Z}^+ .

    \rm (ⅲ) For any prime p>5 , we have

    \begin{equation} \begin{aligned}& \left( \frac{-1}p \right)\sum\limits_{k = 0}^{p-1} \frac{f_kT_k(8,1)}{480^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,4\ ({\rm{mod}}\ {15})\ &\ p = x^2+15y^2\ (x,y\in{\Bbb Z}), \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}2,8\ ({\rm{mod}}\ {15})\ &\ p = 3x^2+5y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-15}p) = -1.\end{cases} \end{aligned} \end{equation} (6.16)

    Remark 6.1 This conjecture was formulated by the author on Oct. 25, 2019.

    Conjecture 6.3. For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1{4n}\sum\limits_{k = 0}^{n-1}(-1)^{n-1-k}(105k+88)f_kT_k(8,1)\in{\Bbb Z}^+. \end{equation} (6.17)

    \rm (ⅱ) Let p be an odd prime. Then

    \begin{equation} \sum\limits_{k = 0}^{p-1}(-1)^k(105k+88)f_kT_k(8,1) {\equiv} \frac 83p \left(23 \left( \frac {-3}p \right)+10 \left( \frac{15}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (6.18)

    If ( \frac{-5}p) = 1 , then

    \begin{equation} \sum\limits_{k = 0}^{pn-1}(-1)^k(105k+88)f_kT_k(8,1)-p \left( \frac p3 \right)\sum\limits_{k = 0}^{n-1}(-1)^k(105k+88)f_kT_k(8,1) \end{equation} (6.19)

    divided by (pn)^2 is a p -adic integer for any n\in{\Bbb Z}^+ .

    \rm (ⅲ) Let p>5 be a prime. Then

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1}(-1)^kf_kT_k(8,1) \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,4\ ({\rm{mod}}\ {15})\ &\ p = x^2+15y^2\ (x,y\in{\Bbb Z}), \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}2,8\ ({\rm{mod}}\ {15})\ &\ p = 3x^2+5y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-15}p) = -1.\end{cases} \end{aligned} \end{equation} (6.20)

    Remark 6.2. This conjecture is the dual of Conjecture 6.2.

    The following conjecture is related to the identity (6.8) .

    Conjecture 6.4. \rm(ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1{2n}\sum\limits_{k = 0}^{n-1}(-1)^k(585k+172)13552^{n-1-k}f_kT_k(1454,1)\in{\Bbb Z}^+. \end{equation} (6.21)

    \rm (ⅱ) Let p>2 be a prime with p\not = 7,11 . Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{585k+172}{(-13552)^k}f_kT_k(1454,1){\equiv} \frac p{11} \left(1580 \left( \frac{-7}p \right) +312 \left( \frac{273}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (6.22)

    If ( \frac {-39}p) = 1 , then

    \begin{equation} \sum\limits_{k = 0}^{pn-1} \frac{585k+172}{(-13552)^k}f_kT_k(1454,1) -p \left( \frac p7 \right)\sum\limits_{k = 0}^{n-1} \frac{585k+172}{(-13552)^k}f_kT_k(1454,1) \end{equation} (6.23)

    divided by (pn)^2 is a p -adic integer for any n\in{\Bbb Z}^+ .

    \rm (ⅲ) Let p>3 be a prime with p\not = 7,11,13 . Then

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{f_kT_k(1454,1)}{(-13552)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p7) = ( \frac p{13}) = 1,\ p = x^2+273y^2, \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{7}) = 1,\ ( \frac p3) = ( \frac p{13}) = -1,\ 2p = x^2+273y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p7) = -1,\ ( \frac p3) = ( \frac p{13}) = 1,\ p = 3x^2+91y^2, \\2p-6x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p7) = ( \frac p{13}) = -1,\ 2p = 3x^2+91y^2, \\28x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{13}) = -1,\ ( \frac p3) = ( \frac p{7}) = 1,\ p = 7x^2+39y^2, \\14x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = -1,\ ( \frac p7) = ( \frac p{13}) = 1,\ 2p = 7x^2+39y^2, \\52x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{3}) = 1,\ ( \frac p7) = ( \frac p{13}) = -1,\ p = 13x^2+21y^2, \\26x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{13}) = 1,\ ( \frac p3) = ( \frac p{7}) = -1,\ 2p = 13x^2+21y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-273}p) = -1, \end{cases} \end{aligned} \end{equation} (6.24)

    where x and y are integers.

    Remark 6.3. Note that the imaginary quadratic field {\Bbb Q}(\sqrt{-273}) has class number 8 .

    The following conjecture is related to the identity (6.10) .

    Conjecture 6.5. \rm(ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1{2n}\sum\limits_{k = 0}^{n-1}(-1)^k(94185k+17014)105984^{n-1-k}f_kT_k(2302,1)\in{\Bbb Z}^+. \end{equation} (6.25)

    \rm (ⅱ) Let p>3 be a prime with p\not = 23 . Then

    \begin{equation} \begin{aligned} &\sum\limits_{k = 0}^{p-1} \frac{94185k+17014}{(-105984)^k}f_kT_k(2302,1) \\{\equiv}& \frac p{16} \left(22659+249565 \left( \frac{-23}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{aligned} \end{equation} (6.26)

    If ( \frac p{23}) = 1 , then

    \begin{equation} \sum\limits_{k = 0}^{pn-1} \frac{94185k+17014}{(-105984)^k}f_kT_k(2302,1) -p\sum\limits_{k = 0}^{n-1} \frac{94185k+17014}{(-105984)^k}f_kT_k(2302,1) \end{equation} (6.27)

    divided by (pn)^2 is a p -adic integer for any n\in{\Bbb Z}^+ .

    \rm (ⅲ) Let p>3 be a prime with p\not = 23 . Then

    \begin{equation} \begin{aligned} &\sum\limits_{k = 0}^{p-1} \frac{f_kT_k(2302,1)}{(-105984)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p5) = ( \frac p{23}) = 1,\ p = x^2+345y^2, \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{23}) = 1,\ ( \frac p3) = ( \frac p{5}) = -1,\ 2p = x^2+345y^2, \\12x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p5) = -1,\ ( \frac p3) = ( \frac p{23}) = 1,\ p = 3x^2+115y^2, \\6x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = -1,\ ( \frac p5) = ( \frac p{23}) = 1,\ 2p = 3x^2+115y^2, \\2p-20x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{5}) = 1,\ ( \frac p3) = ( \frac p{23}) = -1,\ p = 5x^2+69y^2, \\2p-10x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = 1,\ ( \frac p5) = ( \frac p{23}) = -1,\ 2p = 5x^2+69y^2, \\2p-60x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{3}) = ( \frac p5) = ( \frac p{23}) = -1,\ p = 15x^2+23y^2, \\2p-30x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{23}) = -1,\ ( \frac p3) = ( \frac p{5}) = 1,\ 2p = 15x^2+23y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-345}p) = -1, \end{cases} \end{aligned} \end{equation} (6.28)

    where x and y are integers.

    Remark 6.4. Note that the imaginary quadratic field {\Bbb Q}(\sqrt{-345}) has class number 8 .

    The following conjecture is related to the identity (6.12) .

    Conjecture 6.6. \rm(ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1{n}\sum\limits_{k = 0}^{n-1}(210k+23)475904^{n-1-k}f_kT_k(16898,1)\in{\Bbb Z}^+. \end{equation} (6.29)

    \rm (ⅱ) Let p be an odd prime with p\not = 11,13 . Then

    \begin{equation} \begin{aligned} &\sum\limits_{k = 0}^{p-1} \frac{210k+23}{475904^k}f_kT_k(16898,1) \\{\equiv}& \frac p{1287} \left(40621 \left( \frac{-231}p \right)-11020 \left( \frac{66}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{aligned} \end{equation} (6.30)

    If ( \frac {-14}p) = 1 , then

    \begin{equation} \begin{aligned} &\sum\limits_{k = 0}^{pn-1} \frac{210k+23}{475904^k}f_kT_k(16898,1) \\&-p \left( \frac{66}p \right)\sum\limits_{k = 0}^{n-1} \frac{210k+23}{475904^k}f_kT_k(16898,1) \end{aligned} \end{equation} (6.31)

    divided by (pn)^2 is a p -adic integer for any n\in{\Bbb Z}^+ .

    \rm (ⅲ) Let p>3 be a prime with p\not = 7,11,13 . Then

    \begin{equation} \begin{aligned} &\sum\limits_{k = 0}^{p-1} \frac{f_kT_k(16898,1)}{475904^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p3) = ( \frac p7) = ( \frac p{11}) = 1\ &\ p = x^2+462y^2, \\8x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p7) = 1,\ ( \frac p3) = ( \frac p{11}) = -1\ &\ p = 2x^2+231y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p7) = -1,\ ( \frac p3) = ( \frac p{11}) = 1\ &\ p = 3x^2+154y^2, \\2p-24x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p3) = ( \frac p7) = ( \frac p{11}) = -1\ &\ p = 6x^2+77y^2, \\28x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p{3}) = 1,\ ( \frac p7) = ( \frac p{11}) = -1\ &\ p = 7x^2+66y^2, \\44x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p3) = -1,\ ( \frac p7) = ( \frac p{11}) = 1\ &\ p = 11x^2+42y^2, \\56x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p{11}) = 1,\ ( \frac p3) = ( \frac p7) = -1\ &\ p = 14x^2+33y^2, \\2p-84x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p{11}) = -1,\ ( \frac p3) = ( \frac p{7}) = 1\ &\ p = 21x^2+22y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-462}p) = -1, \end{cases} \end{aligned} \end{equation} (6.32)

    where x and y are integers.

    Remark 6.5. Note that the imaginary quadratic field {\Bbb Q}(\sqrt{-462}) has class number 8 .

    The identities (6.5),\, (6.6),\,(6.7),\,(6.9),\,(6.11) are related to the quadratic fields

    {\Bbb Q}(\sqrt{-165}),\ {\Bbb Q}(\sqrt{-210}),\ {\Bbb Q}(\sqrt{-210}),\ {\Bbb Q}(\sqrt{-330}),\ {\Bbb Q}(\sqrt{-357})

    (with class number 8 ) respectively. We also have conjectures on related congruences similar to Conjectures 6.4, 6.5 and 6.6.

    For n\in{\Bbb N} let

    g_n: = \sum\limits_{k = 0}^n \binom nk^2 \binom{2k}k.

    It is known that g_n = \sum_{k = 0}^n \binom nk f_k for all n\in{\Bbb N} . See [43,20,26] for some congruences on polynomials related to these numbers.

    Let p>3 be a prime. For any k = 0,\ldots,p-1 , we have

    g_k{\equiv} \left( \frac{-3}p \right)9^kg_{p-1-k}\ ({\rm{mod}}\ p)

    by [24,Lemma 2.7(ⅱ)]. Combining this with Remark 1.3(ⅱ), we see that

    \begin{align*} \sum\limits_{k = 0}^{p-1} \frac{g_kT_k(b,c)}{m^k}{\equiv}& \left( \frac{-3(b^2-4c)}p \right)\sum\limits_{k = 0}^{p-1} \left( \frac{9(b^2-4c)}m \right)^kg_{p-1-k}T_{p-1-k}(b,c) \\{\equiv}& \left( \frac{3(4c-b^2)}p \right)\sum\limits_{k = 0}^{p-1} \frac{g_kT_k(b,c)}{(9(b^2-4c)/m)^k}\ ({\rm{mod}}\ p) \end{align*}

    for any b,c,m\in{\Bbb Z} with p\nmid (b^2-4c)m .

    Wan and Zudilin [49] obtained the following irrational series for 1/\pi involving the Legendre polynomials and the sequence (g_n)_{n{\geq}0} :

    \sum\limits_{k = 0}^\infty(22k+7-3\sqrt3)g_kP_k \left( \frac{\sqrt{14\sqrt3-15}}3 \right) \left( \frac{\sqrt{2\sqrt3-3}}{9} \right)^k = \frac{9}{2\pi}(9+4\sqrt3).

    Using our congruence approach (including Conjecture 1.4), we find 12 rational series for 1/\pi involving T_n(b,c) and g_n ; Theorem 1 of [49] might be helpful to solve some of them.

    Conjecture 7.1. We have the following identities.

    \begin{align} \sum\limits_{k = 0}^\infty \frac{8k+3}{(-81)^k}g_kT_k(7,-8)& = \frac{9\sqrt3}{4\pi}, \end{align} (7.1)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{4k+1}{(-1089)^k}g_kT_k(31,-32)& = \frac{33}{16\pi}, \end{align} (7.2)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{7k-1}{540^k}g_kT_k(52,1)& = \frac{30\sqrt3}{\pi}, \end{align} (7.3)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{20k+3}{3969^k}g_kT_k(65,64)& = \frac{63\sqrt3}{8\pi}, \end{align} (7.4)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{280k+93}{(-1980)^k}g_kT_k(178,1)& = \frac{20\sqrt{33}}{\pi}, \end{align} (7.5)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{176k+15}{12600^k}g_kT_k(502,1)& = \frac{25\sqrt{42}}{\pi}, \end{align} (7.6)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{560k-23}{13068^k}g_kT_k(970,1)& = \frac{693\sqrt3}{\pi}, \end{align} (7.7)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{12880k+1353}{105840^k}g_kT_k(2158,1)& = \frac{4410\sqrt3}{\pi}, \end{align} (7.8)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{299k+59}{(-101430)^k}g_kT_k(2252,1)& = \frac{735\sqrt{115}}{64\pi}, \end{align} (7.9)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{385k+118}{(-53550)^k}g_kT_k(4048,1)& = \frac{2415\sqrt{17}}{64\pi}, \end{align} (7.10)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{385k-114}{114264^k}g_kT_k(10582,1)& = \frac{15939\sqrt3}{16\pi}, \end{align} (7.11)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{16016k+1273}{510300^k}g_kT_k(17498,1)& = \frac{14175\sqrt3}{2\pi}. \end{align} (7.12)

    Now we present a conjecture on congruences related to (7.6) .

    Conjecture 7.2. \rm(ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1{3n}\sum\limits_{k = 0}^{n-1}(176k+15)12600^{n-1-k}g_kT_k(502,1)\in{\Bbb Z}^+, \end{equation} (7.13)

    and this number is odd if and only if n\in\{2^a:\ a\in{\Bbb N}\} .

    \rm(ⅱ) Let p>7 be a prime. Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{176k+15}{12600^k}g_kT_k(502,1){\equiv} p \left(26 \left( \frac{-42}p \right)-11 \left( \frac{21}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (7.14)

    If p{\equiv}1,3\ ({\rm{mod}}\ 8) , then

    \begin{equation} \sum\limits_{k = 0}^{pn-1} \frac{176k+15}{12600^k}g_kT_k(502,1) -p \left( \frac{21}p \right)\sum\limits_{k = 0}^{n-1} \frac{176k+15}{12600^k}g_kT_k(502,1) \end{equation} (7.15)

    divided by (pn)^2 is a p -adic integer for any n\in{\Bbb Z}^+ .

    \rm(ⅲ) Let p>7 be a prime. Then

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{g_kT_k(502,1)}{12600^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-2}p) = ( \frac p3) = ( \frac p5) = ( \frac p7) = 1\ &\ p = x^2+210y^2, \\2p-8x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-2}p) = ( \frac p7) = 1,\ ( \frac p3) = ( \frac p5) = -1\ &\ p = 2x^2+105y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-2}p) = ( \frac p3) = 1,\ ( \frac p5) = ( \frac p7) = -1\ &\ p = 3x^2+70y^2, \\2p-20x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-2}p) = ( \frac p3) = ( \frac p5) = ( \frac p7) = -1\ &\ p = 5x^2+42y^2, \\24x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-2}p) = ( \frac p5) = 1,\ ( \frac p3) = ( \frac p7) = -1\ &\ p = 6x^2+35y^2, \\28x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-2}p) = ( \frac p5) = -1,\ ( \frac p3) = ( \frac p7) = 1\ &\ p = 7x^2+30y^2, \\40x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-2}p) = ( \frac p7) = -1,\ ( \frac p3) = ( \frac p5) = 1\ &\ p = 10x^2+21y^2, \\56x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-2}p) = ( \frac p3) = -1,\ ( \frac p5) = ( \frac p7) = 1\ &\ p = 14x^2+15y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-210}p) = -1, \end{cases} \end{aligned} \end{equation} (7.16)

    where x and y are integers.

    Remark 7.1. Note that the imaginary quadratic field {\Bbb Q}(\sqrt{-210}) has class number 8 .

    The following conjecture is related to the identity (7.8) .

    Conjecture 7.3. \rm(ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1{3n}\sum\limits_{k = 0}^{n-1}(12880k+1353)105840^{n-1-k}g_kT_k(2158,1)\in{\Bbb Z}^+, \end{equation} (7.17)

    and this number is odd if and only if n\in\{2^a:\ a\in{\Bbb N}\} .

    \rm(ⅱ) Let p>7 be a prime. Then

    \begin{equation} \begin{aligned} &\sum\limits_{k = 0}^{p-1} \frac{12880k+1353}{105840^k}g_kT_k(2158,1) \\{\equiv}& \frac p{2} \left(3419 \left( \frac {-3}p \right)-713 \left( \frac{5}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{aligned} \end{equation} (7.18)

    If ( \frac p3) = ( \frac p{5}) , then

    \begin{equation} \sum\limits_{k = 0}^{pn-1} \frac{12880k+1353}{105840^k}g_kT_k(2158,1) -p \left( \frac p3 \right)\sum\limits_{k = 0}^{n-1} \frac{12880k+1353}{105840^k}g_kT_k(2158,1) \end{equation} (7.19)

    divided by (pn)^2 is a p -adic integer for any n\in{\Bbb Z}^+ .

    \rm(ⅲ) Let p>11 be a prime. Then

    \begin{equation} \begin{aligned} &\sum\limits_{k = 0}^{p-1} \frac{g_kT_k(2158,1)}{105840^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p5) = ( \frac p{11}) = 1,\ p = x^2+330y^2, \\2p-8x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p5) = ( \frac p{11}) = -1,\ p = 2x^2+165y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{11}) = 1,\ ( \frac p3) = ( \frac p5) = -1,\ p = 3x^2+110y^2, \\2p-20x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = -1,\ ( \frac p5) = ( \frac p{11}) = 1,\ p = 5x^2+66y^2, \\24x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{11}) = -1,\ ( \frac p3) = ( \frac p5) = 1,\ p = 6x^2+55y^2, \\40x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = 1,\ ( \frac p5) = ( \frac p{11}) = -1,\ p = 10x^2+33y^2, \\44x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{5}) = 1,\ ( \frac p3) = ( \frac p{11}) = -1,\ p = 11x^2+30y^2, \\60x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p5) = -1,\ ( \frac p3) = ( \frac p{11}) = 1,\ p = 15x^2+22y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-330}p) = -1, \end{cases} \end{aligned} \end{equation} (7.20)

    where x and y are integers.

    Remark 7.2. Note that the imaginary quadratic field {\Bbb Q}(\sqrt{-330}) has class number 8 .

    Now we pose a conjecture related to the identity (7.10) .

    Conjecture 7.4. \rm(ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1{2n}\sum\limits_{k = 0}^{n-1}(-1)^k(385k+118)53550^{n-1-k}g_kT_k(4048,1)\in{\Bbb Z}^+. \end{equation} (7.21)

    \rm(ⅱ) Let p>7 be a prime with p\not = 17 . Then

    \begin{equation} \begin{aligned} &\sum\limits_{k = 0}^{p-1} \frac{385k+118}{(-53550)^k}g_kT_k(4048,1) \\{\equiv}& \frac p{320} \left(29279 \left( \frac{-17}p \right)+8481 \left( \frac{7}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{aligned} \end{equation} (7.22)

    If ( \frac p7) = ( \frac p{17}) , then

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{385k+118}{(-53550)^k}g_kT_k(4048,1) -p \left( \frac{7}p \right)\sum\limits_{k = 0}^{n-1} \frac{385k+118}{(-53550)^k}g_kT_k(4048,1)\right) \end{equation} (7.23)

    is a p -adic integer for any n\in{\Bbb Z}^+ .

    \rm(ⅲ) Let p>7 be a prime with p\not = 17 . Then

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{g_kT_k(4048,1)}{(-53550)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p7) = ( \frac p{17}) = 1,\ p = x^2+357y^2, \\2p-2x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = -1,\ ( \frac p7) = ( \frac p{17}) = 1,\ 2p = x^2+357y^2, \\12x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p5) = ( \frac p7) = -1,\ p = 3x^2+119y^2, \\2p-6x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = 1,\ ( \frac p7) = ( \frac p{17}) = -1,\ 2p = 3x^2+119y^2, \\28x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{17}) = -1,\ ( \frac p3) = ( \frac p7) = 1,\ p = 7x^2+51y^2, \\2p-14x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p7) = 1,\ ( \frac p3) = ( \frac p{17}) = -1,\ 2p = 7x^2+51y^2, \\2p-68x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{17}) = 1,\ ( \frac p3) = ( \frac p7) = -1,\ p = 17x^2+21y^2, \\34x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p7) = -1,\ ( \frac p3) = ( \frac p{17}) = 1,\ 2p = 17x^2+21y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-357}p) = -1, \end{cases} \end{aligned} \end{equation} (7.24)

    where x and y are integers.

    Remark 7.3. Note that the imaginary quadratic field {\Bbb Q}(\sqrt{-357}) has class number 8 .

    Now we pose a conjecture related to the identity (7.12) .

    Conjecture 7.5. \rm(ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1{n}\sum\limits_{k = 0}^{n-1}(16016k+1273)510300^{n-1-k}g_kT_k(17498,1)\in{\Bbb Z}^+, \end{equation} (7.25)

    and this number is odd if and only if n\in\{2^a:\ a\in{\Bbb N}\} .

    \rm(ⅱ) Let p>7 be a prime. Then

    \begin{equation} \begin{aligned} &\sum\limits_{k = 0}^{p-1} \frac{16016k+1273}{510300^k}g_kT_k(17498,1) \\{\equiv}& \frac p{3} \left(6527 \left( \frac{-3}p \right)-2708 \left( \frac{42}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{aligned} \end{equation} (7.26)

    If ( \frac {-14}p) = 1 , then

    \begin{equation} \begin{aligned} &\sum\limits_{k = 0}^{pn-1} \frac{16016k+1273}{510300^k}g_kT_k(17498,1) \\&-p \left( \frac{p}3 \right)\sum\limits_{k = 0}^{n-1} \frac{16016k+1273}{510300^k}g_kT_k(17498,1) \end{aligned} \end{equation} (7.27)

    divided by (pn)^2 is a p -adic integer for each n\in{\Bbb Z}^+ .

    \rm(ⅲ) Let p>11 be a prime. Then

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{g_kT_k(17498,1)}{510300^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p3) = ( \frac p7) = ( \frac p{11}) = 1\ &\ p = x^2+462y^2, \\2p-8x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p7) = 1,\ ( \frac p3) = ( \frac p{11}) = -1\ &\ p = 2x^2+231y^2, \\12x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p7) = -1,\ ( \frac p3) = ( \frac p{11}) = 1\ &\ p = 3x^2+154y^2, \\2p-24x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p3) = ( \frac p7) = ( \frac p{11}) = -1\ &\ p = 6x^2+77y^2, \\28x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p{3}) = 1,\ ( \frac p7) = ( \frac p{11}) = -1\ &\ p = 7x^2+66y^2, \\44x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p3) = -1,\ ( \frac p7) = ( \frac p{11}) = 1\ &\ p = 11x^2+42y^2, \\2p-56x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p{11}) = 1,\ ( \frac p3) = ( \frac p7) = -1\ &\ p = 14x^2+33y^2, \\84x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p{11}) = -1,\ ( \frac p3) = ( \frac p{7}) = 1\ &\ p = 21x^2+22y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-462}p) = -1, \end{cases} \end{aligned} \end{equation} (7.28)

    where x and y are integers.

    Remark 7.4. Note that the imaginary quadratic field {\Bbb Q}(\sqrt{-462}) has class number 8 . We believe that 462 is the largest positive squarefree number d for which the imaginary quadratic field {\Bbb Q}(\sqrt{-d}) can be used to construct a Ramanujan-type series for 1/\pi .

    The identities (7.5),\,(7.7),\,(7.9),\,(7.11) are related to the imaginary quadratic fields {\Bbb Q}(\sqrt{-165}) , {\Bbb Q}(\sqrt{-210}) , {\Bbb Q}(\sqrt{-345}) , {\Bbb Q}(\sqrt{-330}) (with class number 8 ) respectively. We also have conjectures on related congruences similar to Conjectures 7.2, 7.3, 7.4 and 7.5.

    To conclude this section, we confirm an open series for 1/\pi conjectured by the author (cf. [34,(3.28)] and [35,Conjecture 7.9]) in 2011.

    Theorem 7.1. We have

    \begin{equation} \sum\limits_{n = 0}^\infty \frac{16n+5}{324^n} \binom{2n}ng_n(-20) = \frac{189}{25\pi}, \end{equation} (7.29)

    where

    g_n(x): = \sum\limits_{k = 0}^n \binom nk^2 \binom{2k}kx^k.

    Proof. The Franel numbers of order 4 are given by f_n^{(4)} = \sum_{k = 0}^n \binom nk^4\ (n\in{\Bbb N}) . Note that

    f_n^{(4)} \leq\left(\sum\limits_{k = 0}^n \binom nk^2\right)^2 = \binom{2n}n^2 \leq ((1+1)^{2n})^2 = 16^n.

    By [11,(8.1)], for |x|<1/16 and a,b\in{\Bbb Z} , we have

    \begin{equation} \begin{aligned}&\sum\limits_{n = 0}^\infty \binom{2n}n \frac{(an+b)x^n}{(1+2x)^{2n}}\sum\limits_{k = 0}^n \binom nk^2 \binom{2(n-k)}{n-k}x^k \\ = &(1+2x)\sum\limits_{n = 0}^\infty \left( \frac{4a(1-x)(1+2x)n+6ax(2-x)}{5(1-4x)}+b \right)f_n^{(4)}x^n. \end{aligned} \end{equation} (7.30)

    Since

    \begin{align*} & \frac{x^n}{(1+2x)^{2n}}\sum\limits_{k = 0}^n \binom nk^2 \binom{2n-2k}{n-k}x^k \\ = & \frac{x^n}{(1+2x)^{2n}}\sum\limits_{k = 0}^n \binom nk^2 \binom{2k}{k}x^{n-k} = (2+x^{-1})^{-2n}g_n(x^{-1}), \end{align*}

    putting a = 16 , b = 5 and x = -1/20 in (7.30) we obtain

    \sum\limits_{n = 0}^\infty \frac{16n+5}{18^{2n}} \binom{2n}ng_n(-20) = \frac{378}{125}\sum\limits_{n = 0}^\infty \frac{3n+1}{(-20)^n}f_n^{(4)}.

    As

    \sum\limits_{n = 0}^\infty \frac{3n+1}{(-20)^n}f_n^{(4)} = \frac{5}{2\pi}

    by Cooper [9], we finally get

    \sum\limits_{n = 0}^\infty \frac{16n+5}{18^{2n}} \binom{2n}ng_n(-20) = \frac{378}{125}\times \frac 5{2\pi} = \frac{189}{25\pi}.

    This concludes the proof of (7.29).

    Recall that the numbers

    \beta_n: = \sum\limits_{k = 0}^n \binom nk^2 \binom{n+k}k\ \ \ (n = 0,1,2,\ldots)

    are a kind of Apéry numbers. Let p be an odd prime. For any k = 0,1,\ldots,p-1 , we have

    \beta_k{\equiv}(-1)^k\beta_{p-1-k}\ ({\rm{mod}}\ p)

    by [24,Lemma 2.7(ⅰ)]. Combining this with Remark 1.3(ⅱ), we see that

    \begin{align*} \sum\limits_{k = 0}^{p-1} \frac{\beta_kT_k(b,c)}{m^k}{\equiv}& \left( \frac{b^2-4c}p \right)\sum\limits_{k = 0}^{p-1} \left( \frac{-(b^2-4c)}m \right)^k\beta_{p-1-k}T_{p-1-k}(b,c) \\{\equiv}& \left( \frac{b^2-4c}p \right)\sum\limits_{k = 0}^{p-1} \frac{\beta_kT_k(b,c)}{((4c-b^2)/m)^k}\ ({\rm{mod}}\ p) \end{align*}

    for any b,c,m\in{\Bbb Z} with p\nmid (b^2-4c)m .

    Wan and Zudilin [49] obtained the following irrational series for 1/\pi involving the Legendre polynomials and the numbers \beta_n :

    \sum\limits_{k = 0}^\infty(60k+16-5\sqrt{10})\beta_kP_k \left( \frac{5\sqrt2+17\sqrt5}{45} \right) \left( \frac{5\sqrt2-3\sqrt5} 5 \right)^k = \frac{135\sqrt2+81\sqrt5}{\sqrt2\,\pi}.

    Using our congruence approach (including Conjecture 1.4), we find one rational series for 1/\pi involving T_n(b,c) and the Apéry numbers \beta_n (see (8.1) below); Theorem 1 of [49] might be helpful to solve it.

    Conjecture 8.1. (ⅰ) We have

    \begin{equation} \sum\limits_{k = 0}^\infty \frac{145k+9}{900^k}\beta_kT_k(52,1) = \frac{285}{\pi}. \end{equation} (8.1)

    Also, for any n\in{\Bbb Z}^+ we have

    \begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(145k+9)900^{n-1-k}\beta_kT_k(52,1)\in{\Bbb Z}^+. \end{equation} (8.2)

    \rm(ⅱ) Let p>5 be a prime. Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{145k+9}{900^k}\beta_kT_k(52,1) {\equiv} \frac p5 \left(133 \left( \frac{-1}p \right)-88 \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (8.3)

    If p{\equiv}1\ ({\rm{mod}}\ 4) , then

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{145k+9}{900^k}\beta_kT_k(52,1) -p\sum\limits_{k = 0}^{n-1} \frac{145k+9}{900^k}\beta_kT_k(52,1)\right) \in{\Bbb Z}_p \end{equation} (8.4)

    for all n\in{\Bbb Z}^+ .

    \rm(ⅲ) Let p>5 be a prime. Then

    \begin{equation} \begin{aligned}& \left( \frac{-1}p \right)\sum\limits_{k = 0}^{p-1} \frac{\beta_kT_k(52,1)}{900^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,4\ ({\rm{mod}}\ {15})\ &\ p = x^2+15y^2\ (x,y\in{\Bbb Z}), \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}2,8\ ({\rm{mod}}\ {15})\ &\ p = 3x^2+5y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-15}p) = -1.\end{cases} \end{aligned} \end{equation} (8.5)

    Remark 8.1. This conjecture was formulated by the author on Oct. 27, 2019.

    Conjecture 8.2. \rm(ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1{2n}\sum\limits_{k = 0}^{n-1}(-1)^k(15k+8)\beta_kT_k(4,-1)\in{\Bbb Z}, \end{equation} (8.6)

    and this number is odd if and only if n\in\{2^a:\ a\in{\Bbb Z}^+\} .

    \rm(ⅱ) Let p be a prime. Then

    \begin{equation} \sum\limits_{k = 0}^{p-1}(-1)^k(15k+8)\beta_kT_k(4,-1) {\equiv} \frac p4 \left(27 \left( \frac p3 \right)+5 \left( \frac p5 \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (8.7)

    If ( \frac {-15}p) = 1\ ( i.e., p{\equiv}1,2,4,8\ ({\rm{mod}}\ {15})) , then

    \begin{equation} \sum\limits_{k = 0}^{pn-1}(-1)^k(15k+8)\beta_kT_k(4,-1) -p \left( \frac p3 \right)\sum\limits_{k = 0}^{n-1}(-1)^k(15k+8)\beta_kT_k(2,2) \end{equation} (8.8)

    divided by (pn)^2 is a p -adic integer for any n\in{\Bbb Z}^+ .

    \rm(ⅲ) For any prime p>5 , we have

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1}(-1)^k\beta_kT_k(4,-1) \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,4\ ({\rm{mod}}\ {15})\ &\ p = x^2+15y^2\ (x,y\in{\Bbb Z}), \\12x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}2,8\ ({\rm{mod}}\ {15})\ &\ p = 3x^2+5y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac {-15}p) = -1.\end{cases} \end{aligned} \end{equation} (8.9)

    Remark 8.2. This conjecture was formulated by the author on Nov. 13, 2019.

    Conjecture 8.3. \rm(ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac3{n2^{\lfloor n/2\rfloor}}\sum\limits_{k = 0}^{n-1}(2k+1)(-2)^{n-1-k}\beta_kT_k(2,2)\in{\Bbb Z}^+, \end{equation} (8.10)

    and this number is odd if and only if n is a power of two.

    \rm(ⅱ) Let p>3 be a prime. Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{2k+1}{(-2)^k}\beta_kT_k(2,2) {\equiv} \frac p3 \left(1+2 \left( \frac{-1}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (8.11)

    If p{\equiv}1\ ({\rm{mod}}\ 4) , then

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{2k+1}{(-2)^k}\beta_kT_k(2,2) -p\sum\limits_{k = 0}^{n-1} \frac{2k+1}{(-2)^k}\beta_kT_k(2,2)\right)\in{\Bbb Z}_p \end{equation} (8.12)

    for all n\in{\Bbb Z}^+ .

    \rm(ⅲ) For any odd prime p , we have

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{\beta_kT_k(2,2)}{(-2)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1\ ({\rm{mod}}\ 4)\ &\ p = x^2+4y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}3\ ({\rm{mod}}\ 4).\end{cases} \end{aligned} \end{equation} (8.13)

    Remark 8.3. This conjecture was formulated by the author on Nov. 13, 2019.

    Conjecture 8.4. \rm(ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1{n2^{\lfloor(n+1)/2\rfloor}}\sum\limits_{k = 0}^{n-1}(3k+2)(-2)^{n-1-k}\beta_kT_k(20,2)\in{\Bbb Z}^+, \end{equation} (8.14)

    and this number is odd if and only if n\in\{2^a:\ a = 0,2,3,4,\ldots\} .

    \rm(ⅱ) Let p be any odd prime. Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{3k+2}{(-2)^k}\beta_kT_k(20,2) {\equiv}2p \left( \frac 2p \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (8.15)

    If p{\equiv}\pm1\ ({\rm{mod}}\ 8) , then

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{3k+2}{(-2)^k}\beta_kT_k(20,2) -p\sum\limits_{k = 0}^{n-1} \frac{3k+2}{(-2)^k}\beta_kT_k(20,2)\right)\in{\Bbb Z}_p \end{equation} (8.16)

    for all n\in{\Bbb Z}^+ .

    \rm(ⅲ) For any odd prime p , we have

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{\beta_kT_k(20,2)}{(-2)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1\ ({\rm{mod}}\ 4)\ &\ p = x^2+4y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}3\ ({\rm{mod}}\ 4).\end{cases} \end{aligned} \end{equation} (8.17)

    Conjecture 8.5. \rm(ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac3n\sum\limits_{k = 0}^{n-1}(5k+3)4^{n-1-k}\beta_kT_k(14,-1)\in{\Bbb Z}. \end{equation} (8.18)

    \rm(ⅱ) Let p>3 be a prime. Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{5k+3}{4^k}\beta_kT_k(14,-1) {\equiv} \frac p3 \left(4 \left( \frac{-2}p \right)+5 \left( \frac 2p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (8.19)

    If p{\equiv}1\ ({\rm{mod}}\ 4) , then

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{5k+3}{4^k}\beta_kT_k(14,-1) -p \left( \frac 2p \right)\sum\limits_{k = 0}^{n-1} \frac{5k+3}{4^k}\beta_kT_k(14,-1)\right)\in{\Bbb Z}_p \end{equation} (8.20)

    for all n\in{\Bbb Z}^+ .

    \rm(ⅲ) Let p\not = 2,5 be a prime. Then

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{\beta_kT_k(14,-1)}{4^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac {-2}p) = ( \frac 5p) = 1\ &\ p = x^2+10y^2\ (x,y\in{\Bbb Z}), \\8x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac {-2}p) = ( \frac 5p) = -1\ &\ p = 2x^2+5y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac {-10}p) = -1.\end{cases} \end{aligned} \end{equation} (8.21)

    Conjecture 8.6. \rm(ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1{3n}\sum\limits_{k = 0}^{n-1}(22k+15)(-4)^{n-1-k}\beta_kT_k(46,1)\in{\Bbb Z}^+, \end{equation} (8.22)

    and this number is odd if and only if n is a power of two.

    \rm(ⅱ) Let p be an odd prime. Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{22k+15}{(-4)^k}\beta_kT_k(46,1) {\equiv} \frac p4 \left(357-297 \left( \frac{33}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (8.23)

    If ( \frac{33}p) = 1 , then

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{22k+15}{(-4)^k}\beta_kT_k(46,1) -p\sum\limits_{k = 0}^{n-1} \frac{22k+15}{(-4)^k}\beta_kT_k(46,1)\right)\in{\Bbb Z}_p \end{equation} (8.24)

    for all n\in{\Bbb Z}^+ .

    \rm(ⅲ) Let p>3 be a prime. Then

    \begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{\beta_kT_k(46,1)}{(-4)^k} \\{\equiv}&\begin{cases}x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p{11}) = 1\ &\ 4p = x^2+11y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p{11}) = -1.\end{cases} \end{aligned} \end{equation} (8.25)

    Conjecture 8.7. \rm(ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(190k+91)(-60)^{n-1-k}\beta_kT_k(82,1)\in{\Bbb Z}^+, \end{equation} (8.26)

    and this number is odd if and only if n is a power of two.

    \rm(ⅱ) Let p>5 be a prime. Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{190k+91}{(-60)^k}\beta_kT_k(82,1) {\equiv} \frac p4 \left(111+253 \left( \frac{-15}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (8.27)

    If ( \frac{-15}p) = 1 , then

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{190k+91}{(-60)^k}\beta_kT_k(82,1) -p\sum\limits_{k = 0}^{n-1} \frac{190k+91}{(-60)^k}\beta_kT_k(82,1)\right)\in{\Bbb Z}_p \end{equation} (8.28)

    for all n\in{\Bbb Z}^+ .

    \rm(ⅲ) For any prime p>7 , we have

    \begin{equation} \begin{aligned}& \left( \frac p3 \right)\sum\limits_{k = 0}^{p-1} \frac{\beta_kT_k(82,1)}{(-60)^k} \\{\equiv}&\begin{cases}x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p{5}) = ( \frac p7) = 1\ &\ 4p = x^2+35y^2\ (x,y\in{\Bbb Z}), \\2p-5x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p{5}) = ( \frac p7) = -1\ &\ 4p = 5x^2+7y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac {-35}p) = -1.\end{cases} \end{aligned} \end{equation} (8.29)

    The numbers

    w_n: = \sum\limits_{k = 0}^{\lfloor n/3\rfloor}(-1)^k3^{n-3k} \binom n{3k} \binom{3k}k \binom{2k}k\ \ \ (n = 0,1,2,\ldots)

    were first introduced by Zagier [51] during his study of Apéry-like integer sequences, who noted the recurrence

    (n+1)^2w_{n+1} = (9n(n+1)+3)w_n-27n^2w_{n-1}\ (n = 1,2,3,\ldots).

    Lemma 9.1. Let p>3 be a prime. Then

    w_k{\equiv} \left( \frac{-3}p \right)27^kw_{p-1-k}\ ({\rm{mod}}\ p)\quad \mathit{\text{for all}}\ k = 0,\ldots,p-1.

    Proof. Note that

    \begin{align*} w_{p-1} = &\sum\limits_{k = 0}^{\lfloor(p-1)/3\rfloor}(-1)^k3^{p-1-3k} \binom{p-1}{3k} \binom{3k}k \binom{2k}k \\{\equiv}&\sum\limits_{k = 0}^{p-1} \frac{ \binom{2k}k \binom{3k}k}{27^k}{\equiv} \left( \frac p3 \right)\ ({\rm{mod}}\ p) \end{align*}

    with the help of the known congruence \sum_{k = 0}^{p-1} \binom{2k}k \binom{3k}k/27^k{\equiv}( \frac p3)\ ({\rm{mod}}\ {p^2}) conjectured by F. Rodriguez-Villegas [28] and proved by E. Mortenson [25]. Similarly,

    \begin{align*} w_{p-2} = &\sum\limits_{k = 0}^{\lfloor(p-2)/3\rfloor}(-1)^k3^{p-2-3k} \binom{p-2}{3k} \binom{3k}k \binom{2k}k \\ = &\sum\limits_{k = 0}^{\lfloor(p-2)/3\rfloor}(-1)^k3^{p-2-3k} \frac{3k+1}{p-1} \binom{p-1}{3k+1} \binom{3k}k \binom{2k}k \\{\equiv}& \frac19\sum\limits_{k = 0}^{p-1}(9k+3) \frac{ \binom{2k}k \binom{3k}k}{27^k} {\equiv} \frac19 \left( \frac p3 \right)+ \frac19\sum\limits_{k = 0}^{p-1}(9k+2) \frac{ \binom{2k}k \binom{3k}k}{27^k}\ ({\rm{mod}}\ p). \end{align*}

    By induction,

    \sum\limits_{k = 0}^n(9k+2) \frac{ \binom{2k}k \binom{3k}k}{27^k} = (3n+1)(3n+2) \frac{ \binom{2n}n \binom{3n}n}{27^n}

    for all n\in{\Bbb N} . In particular,

    \sum\limits_{k = 0}^{p-1}(9k+2) \frac{ \binom{2k}k \binom{3k}k}{27^k} = \frac{(3p-2)(3p-1)}{27^{p-1}}pC_{p-1} \binom{3p-3}{p-1}{\equiv}0\ ({\rm{mod}}\ p).

    So we have w_k{\equiv}( \frac{-3}p)27^kw_{p-1-k}\ ({\rm{mod}}\ p) for k = 0,1 . (Note that w_0 = 1 and w_1 = 3 .)

    Now let k\in\{1,\ldots,p-2\} and assume that

    w_j{\equiv} \left( \frac{-3}p \right)27^jw_{p-1-j}\quad\text{for all}\ j = 0,\ldots,k.

    Then

    \begin{align*} &(k+1)^2w_{k+1} = (9k(k+1)+3)w_k-27k^2w_{k-1} \\{\equiv}&(9(p-k)(p-k-1)+3) \left( \frac{-3}p \right)27^kw_{p-1-k} -27(p-k)^2 \left( \frac{-3}p \right)27^{k-1}w_{p-1-(k-1)} \\ = & \left( \frac{-3}p \right)27^k\times 27(p-k-1)^2w_{p-k-2}\ ({\rm{mod}}\ p) \end{align*}

    and hence

    w_{k+1}{\equiv} \left( \frac{-3}p \right)27^{k+1}w_{p-1-(k+1)}\ ({\rm{mod}}\ p).

    In view of the above, we have proved the desired result by induction.

    For Lemma 9.1 one may also consult [31,Corollary 3.1]. Let p>3 be a prime. In view of Lemma 9.1 and Remark 1.3(ⅱ), we have

    \begin{align*} \sum\limits_{k = 0}^{p-1} \frac{w_kT_k(b,c)}{m^k}{\equiv}& \left( \frac{-3(b^2-4c)}p \right)\sum\limits_{k = 0}^{p-1} \left( \frac{27(b^2-4c)}m \right)^kw_{p-1-k}T_{p-1-k}(b,c) \\{\equiv}& \left( \frac{-3(b^2-4c)}p \right)\sum\limits_{k = 0}^{p-1} \frac{w_kT_k(b,c)}{(27(b^2-4c)/m)^k}\ ({\rm{mod}}\ p) \end{align*}

    for any b,c,m\in{\Bbb Z} with p\nmid (b^2-4c)m .

    Wan and Zudilin [49] obtained the following irrational series for 1/\pi involving the Legendre polynomials and the numbers w_n :

    \sum\limits_{k = 0}^\infty(14k+7-\sqrt{21})w_kP_k \left( \frac{\sqrt{21}}{5} \right) \left( \frac{7\sqrt{21}-27} {90} \right)^k = \frac{5\sqrt{7(7\sqrt{21}+27)}}{4\sqrt2\,\pi}.

    Using our congruence approach (including Conjecture 1.4), we find five rational series for 1/\pi involving T_n(b,c) and the numbers w_n ; Theorem 1 of [49] might be helpful to solve them.

    Conjecture 9.1. We have

    \begin{align} \sum\limits_{k = 0}^\infty \frac{13k+3}{100^k}w_kT_k(14,-1)& = \frac{30\sqrt2}{\pi}, \end{align} (9.1)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{14k+5}{108^k}w_kT_k(18,1)& = \frac{27\sqrt3}{\pi}, \end{align} (9.2)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{19k+2}{486^k}w_kT_k(44,-2)& = \frac{81\sqrt3}{4\pi}, \end{align} (9.3)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{91k+32}{(-675)^k}w_kT_k(52,1)& = \frac{45\sqrt3}{2\pi}, \end{align} (9.4)
    \begin{align} \sum\limits_{k = 0}^\infty \frac{182k+37}{756^k}w_kT_k(110,1)& = \frac{315\sqrt3}{\pi}. \end{align} (9.5)

    Below we present our conjectures on congruences related to the identities (9.2) and (9.5).

    Conjecture 9.2. \rm(ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(14k+5)108^{n-1-k}w_kT_k(18,1)\in{\Bbb Z}^+, \end{equation} (9.6)

    and this number is odd if and only if n\in\{2^a:\ a\in{\Bbb N}\} .

    \rm(ⅱ) Let p>3 be a prime. Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{14k+5}{108^k}w_kT_k(18,1){\equiv} \frac p4 \left(27 \left( \frac {-3}p \right)-7 \left( \frac {21}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (9.7)

    If ( \frac p7) = 1\ ( i.e., p{\equiv}1,2,4\ ({\rm{mod}}\ 7)) , then

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{14k+5}{108^k}w_kT_k(18,1)- \left( \frac p3 \right)\sum\limits_{k = 0}^{n-1} \frac{14k+5}{108^k}w_kT_k(18,1)\right)\in{\Bbb Z}_p \end{equation} (9.8)

    for all n\in{\Bbb Z}^+ .

    \rm(ⅲ) For any prime p>7 , we have

    \begin{equation} \begin{aligned}& \left( \frac p3 \right)\sum\limits_{k = 0}^{p-1} \frac{w_kT_k(18,1)}{108^k} \\{\equiv}&\begin{cases}x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p5) = ( \frac p7) = 1\ &\ 4p = x^2+35y^2\ (x,y\in{\Bbb Z}), \\5x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p5) = ( \frac p7) = -1\ &\ 4p = 5x^2+7y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-35}p) = -1. \end{cases}\end{aligned} \end{equation} (9.9)

    Conjecture 9.3. \rm(ⅰ) For any n\in{\Bbb Z}^+ , we have

    \begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(182k+37)756^{n-1-k}w_kT_k(110,1)\in{\Bbb Z}^+, \end{equation} (9.10)

    and this number is odd if and only if n\in\{2^a:\ a\in{\Bbb N}\} .

    \rm(ⅱ) Let p>3 be a prime with p\not = 7 . Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{182k+37}{756^k}w_kT_k(110,1){\equiv} \frac p4 \left(265 \left( \frac {-3}p \right)-117 \left( \frac {21}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (9.11)

    If ( \frac p7) = 1\ ( i.e., p{\equiv}1,2,4\ ({\rm{mod}}\ 7)) , then

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{182k+37}{756^k}w_kT_k(110,1)- \left( \frac p3 \right)\sum\limits_{k = 0}^{n-1} \frac{182k+37}{756^k}w_kT_k(110,1)\right)\in{\Bbb Z}_p \end{equation} (9.12)

    for all n\in{\Bbb Z}^+ .

    \rm(ⅲ) For any prime p>3 with p\not = 7,13 , we have

    \begin{equation} \begin{aligned}& \left( \frac p3 \right)\sum\limits_{k = 0}^{p-1} \frac{w_kT_k(110,1)}{756^k} \\{\equiv}&\begin{cases}x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p7) = ( \frac p{13}) = 1\ &\ 4p = x^2+91y^2\ (x,y\in{\Bbb Z}), \\7x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p7) = ( \frac p{13}) = -1\ &\ 4p = 7x^2+13y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-91}p) = -1. \end{cases}\end{aligned} \end{equation} (9.13)

    Now we give one more conjecture in this section.

    Conjecture 9.4. \rm(ⅰ) For any integer n>1 , we have

    \begin{equation} \frac1{3n2^{\lfloor(n+1)/2\rfloor}} \sum\limits_{k = 0}^{n-1}(2k+1)54^{n-1-k}w_kT_k(10,-2)\in{\Bbb Z}^+. \end{equation} (9.14)

    \rm(ⅱ) Let p>3 be a prime. Then

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{2k+1}{54^k}w_kT_k(10,-2){\equiv} p \left( \frac p3 \right)+ \frac{p}2(2^{p-1}-1) \left(5 \left( \frac p3 \right)+3 \left( \frac 3p \right) \right)\ ({\rm{mod}}\ {p^3}). \end{equation} (9.15)

    If p{\equiv}1\ ({\rm{mod}}\ 4) , then

    \begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{2k+1}{54^k}w_kT_k(10,-2)- \left( \frac p3 \right)\sum\limits_{k = 0}^{n-1} \frac{2k+1}{54^k}w_kT_k(10,-2)\right)\in{\Bbb Z}_p \end{equation} (9.16)

    for all n\in{\Bbb Z}^+ .

    \rm(ⅲ) For any prime p>3 , we have

    \begin{equation} \begin{aligned}& \left( \frac p3 \right)\sum\limits_{k = 0}^{p-1} \frac{w_kT_k(10,-2)}{54^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ 4\mid p-1\ &\ p = x^2+4y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}3\ ({\rm{mod}}\ 4). \end{cases}\end{aligned} \end{equation} (9.17)

    Remark 9.1. For primes p>3 with p{\equiv}3\ ({\rm{mod}}\ 4) , in general the congruence (9.16) is not always valid for all n\in{\Bbb Z}^+ . This does not violate Conjecture 1.2 since \lim_{k\to+\infty}|w_kT_k(10,-2)|^{1/k} = \sqrt{27}\times\sqrt{10^2-4(-2)} = 54 . If the series \sum_{k = 0}^\infty \frac{2k+1}{54^k}w_kT_k(10,-2) converges, its value times \pi/\sqrt3 should be a rational number.

    Let p be an odd prime and let a,b,c,d,m\in{\Bbb Z} with m(b^2-4c)\not{\equiv}0\ ({\rm{mod}}\ p) . Then

    \begin{align*} \sum\limits_{k = 1}^{p-1} \frac{a+dk}{m^k} \binom{2k}k^2T_k(b,c) {\equiv}&\sum\limits_{k = 1}^{(p-1)/2} \frac{a+dk}{k^2m^k} \left(k \binom{2k}k \right)^2T_k(b,c) \\{\equiv}&\sum\limits_{k = 1}^{(p-1)/2} \frac{a+dk}{k^2m^k} \left(- \frac{2p}{ \binom{2(p-k)}{p-k}} \right)^2T_k(b,c)\ ({\rm{mod}}\ p) \end{align*}

    with the aid of [33,Lemma 2.1]. Thus

    \begin{align*} &\sum\limits_{k = 1}^{p-1} \frac{a+dk}{m^k} \binom{2k}k^2T_k(b,c) \\{\equiv}&4p^2\sum\limits_{k = 1}^{(p-1)/2} \frac{a+dk}{k^2m^k}\times \frac{T_k(b,c)}{ \binom{2(p-k)}{p-k}^2} \\{\equiv}&4p^2\sum\limits_{p/2 < k < p} \frac{a+d(p-k)}{(p-k)^2m^{p-k}}\times \frac{T_{p-k}(b,c)}{ \binom{2k}k^2} \\{\equiv}&4p^2\sum\limits_{k = 1}^{p-1} \frac{(a-dk)m^{k-1}}{k^2 \binom{2k}k^2} \left( \frac{b^2-4c}p \right)(b^2-4c)^{p-k}T_{p-1-(p-k)}(b,c) \\{\equiv}& \left( \frac{b^2-4c}p \right)4p^2\sum\limits_{k = 1}^{p-1} \frac{(a-dk)T_{k-1}(b,c)}{k^2 \binom{2k}k^2} \left( \frac{m}{b^2-4c} \right)^{k-1} \ ({\rm{mod}}\ p) \end{align*}

    in view of Remark 1.3(ⅱ).

    Let p>3 be a prime. By the above, the author's conjectural congruence (cf. [35,Conjecture 1.3])

    \sum\limits_{k = 0}^{p-1}(105k+44)(-1)^k \binom{2k}k^2T_k{\equiv} p \left(20+24 \left( \frac p3 \right)(2-3^{p-1}) \right)\ ({\rm{mod}}\ {p^3})

    implies that

    p^2\sum\limits_{k = 1}^{p-1} \frac{(105k-44)T_{k-1}}{k^2 \binom{2k}k^23^{k-1}}{\equiv} 11 \left( \frac p3 \right)\ ({\rm{mod}}\ p).

    Motivated by this, we pose the following curious conjecture.

    Conjecture 10.1. We have the following identities:

    \begin{align} \sum\limits_{k = 1}^\infty \frac{(105k-44)T_{k-1}}{k^2 \binom{2k}k^23^{k-1}} = & \frac{5\pi}{\sqrt3}+6\log3, \end{align} (10.1)
    \begin{align} \sum\limits_{k = 2}^\infty \frac{(5k-2)T_{k-1}}{(k-1)k^2 \binom{2k}k^23^{k-1}} = & \frac{21-2\sqrt3\,\pi-9\log3}{12}. \end{align} (10.2)

    Remark 10.1. The two identities were conjectured by the author on Dec. 7, 2019. One can easily check them numerically via \mathsf{Mathematica} as the two series converge fast.

    Now we state our related conjectures on congruences.

    Conjecture 10.2. For any prime p>3 , we have

    \begin{equation} p^2\sum\limits_{k = 1}^{p-1} \frac{(105k-44)T_{k-1}}{k^2 \binom{2k}k^23^{k-1}} {\equiv} 11 \left( \frac p3 \right)+ \frac p2 \left(13-35 \left( \frac p3 \right) \right)\ ({\rm{mod}}\ {p^2}) \end{equation} (10.3)

    and

    \begin{equation} p^2\sum\limits_{k = 2}^{p-1} \frac{(5k-2)T_{k-1}}{(k-1)k^2 \binom{2k}k^23^{k-1}} {\equiv}- \frac12 \left( \frac p3 \right)- \frac p8 \left(7+ \left( \frac p3 \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (10.4)

    Conjecture 10.3. (ⅰ) We have

    \frac1{n \binom{2n}n}\sum\limits_{k = 0}^{n-1}(-1)^{n-1-k}(5k+2) \binom{2k}kC_kT_k\in{\Bbb Z}^+

    for all n\in{\Bbb Z}^+ , and also

    \sum\limits_{k = 0}^{p-1}(-1)^k(5k+2) \binom{2k}kC_kT_k{\equiv} 2p \left(1- \left( \frac p3 \right)(3^p-3) \right)\ ({\rm{mod}}\ {p^3})

    for each prime p>3 .

    \rm(ⅱ) For any prime p{\equiv}1\ ({\rm{mod}}\ 3) and n\in{\Bbb Z}^+ , we have

    \begin{equation} \begin{aligned}& \frac{\sum\limits_{k = 0}^{pn-1}(-1)^k(5k+2) \binom{2k}kC_kT_k-p\sum\limits_{k = 0}^{n-1}(-1)^k(5k+2) \binom{2k}kC_kT_k} {(pn)^2 \binom{2n}n^2} \\\quad\qquad&{\equiv} \left( \frac p3 \right) \frac{3^p-3}{2p}(-1)^nT_{n-1}\ ({\rm{mod}}\ {p}). \end{aligned} \end{equation} (10.5)

    Remark 10.2. See also [45,Conjecture 67] for a similar conjecture.

    Let p be an odd prime. We conjecture that

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{8k+3}{(-16)^k} \binom{2k}k^2T_k(3,-4){\equiv} p \left(1+2 \left( \frac{-1}p \right) \right)\ ({\rm{mod}}\ {p^2}) \end{equation} (10.6)

    and

    \begin{equation} \sum\limits_{k = 0}^{p-1} \frac{33k+14}{4^k} \binom{2k}k^2T_k(8,-2){\equiv} p \left(6 \left( \frac{-1}p \right)+8 \left( \frac{2}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} (10.7)

    Though (10.6) implies the congruence

    p^2\sum\limits_{k = 1}^{p-1} \frac{(8k-3)T_{k-1}(3,-4)}{k^2 \binom{2k}k^2} \left(- \frac{16}{25} \right)^{k-1}{\equiv} \frac 34\ ({\rm{mod}}\ p),

    and (10.7) with p>3 implies the congruence

    p^2\sum\limits_{k = 1}^{p-1} \frac{(33k-14)T_{k-1}(8,-2)}{k^2 \binom{2k}k^218^{k-1}}{\equiv} \frac 7{2} \left( \frac 2p \right)\ ({\rm{mod}}\ p),

    we are unable to find the exact values of the two converging series

    \sum\limits_{k = 1}^\infty \frac{(8k-3)T_{k-1}(3,-4)}{k^2 \binom{2k}k^2} \left(- \frac{16}{25} \right)^{k-1} \ \ \text{and}\ \ \sum\limits_{k = 1}^{\infty} \frac{(33k-14)T_{k-1}(8,-2)}{k^2 \binom{2k}k^218^{k-1}}.

    The author would like to thank Prof. Qing-Hu Hou at Tianjin Univ. for his helpful comments on the proof of Lemma 2.3.



    [1] Sach VJ, Buchner E, Schmieder M (2020) Enigmatic earthquake-generated large-scale clastic dyke in the Biberach area (SW Germany). Sediment Geol 398: 105571. https://doi.org/10.1016/j.sedgeo.2019.105571 doi: 10.1016/j.sedgeo.2019.105571
    [2] Buchner E, Sach VJ, Schmieder M (2020) New discovery of two seismite horizons challenges the Ries–Steinheim double‑impact theory. Sci Rep 10: 22143. https://doi.org/10.1038/s41598-020-79032-4 doi: 10.1038/s41598-020-79032-4
    [3] Stöffler D, Artemieva NA, Wünnemann K, et al. (2013) Ries crater and suevite revisited—Observations and modeling. Part Ⅰ: Observations. Meteorit Planet Sci 48: 515–589. https://doi.org/10.1111/maps.12086 doi: 10.1111/maps.12086
    [4] Artemieva NA, Wünnemann K, Krien F, et al. (2013) Ries crater and suevite revisited—Observations and modeling. Part Ⅱ: Modeling. Meteorit Planet Sci 48: 590–627. https://doi.org/10.1111/maps.12085 doi: 10.1111/maps.12085
    [5] Schmieder M, Kennedy T, Jourdan F, et al. (2018) A high-precision 40Ar/39Ar age for the Nördlinger Ries impact crater, Germany, and implications for the accurate dating of terrestrial impact events. Geochim Cosmochim Acta 220: 146–157. https://doi.org/10.1016/j.gca.2017.09.036 doi: 10.1016/j.gca.2017.09.036
    [6] Schmieder M, Kennedy T, Jourdan F (2018) Response to comment on "A high-precision 40Ar/39Ar age for the Nördlinger Ries impact crater, Germany, and implications for the accurate dating of terrestrial impact events" by Schmieder et al. (Geochim. Cosmochim. Acta 220 (2018) 146–157). Geochim Cosmochim Acta 238: 602–605. https://doi.org/10.1016/j.gca.2018.07.025 doi: 10.1016/j.gca.2018.07.025
    [7] Sach VJ (1999) Litho- und biostratigraphische Untersuchungen in der Oberen Süßwassermolasse des Landkreises Biberach an der Riß (Oberschwaben). Stuttgarter Beitr Naturk B 276: 1–167.
    [8] Sach VJ (2014) Strahlenkalke (Shatter-Cones) aus dem Brockhorizont der Oberen Süßwassermolasse in Oberschwaben (Südwestdeutschland)—Fernauswürflinge des Nördlinger-Ries-Impaktes, Pfeil Verlag, München, 1–17.
    [9] Hofmann B, Hofmann F (1992) An impactite horizon in the upper freshwater molasse in Eastern Switzerland: Distal Ries ejecta. Eclogae Geol Helv 85: 788–789.
    [10] Letsch D (2017) Diamictites and soft sediment deformation related to the Ries (ca. 14.9 Ma) meteorite impact: the "Blockhorizont" of Bernhardzell (Eastern Switzerland). Int J Earth Sci 107: 1379–1380. https://doi.org/10.1007/s00531-017-1542-1 doi: 10.1007/s00531-017-1542-1
    [11] Holm-Alwmark S, Alwmark C, Ferrière L, et al. (2021) Shocked quartz in distal ejecta from the Ries impact event (Germany) found at ~ 180 km distance, near Bernhardzell, eastern Switzerland. Sci Rep 11: 7438. https://doi.org/10.1038/s41598-021-86685-2 doi: 10.1038/s41598-021-86685-2
    [12] Buchner E, Sach VJ, Schmieder M (2022) Event- and biostratigraphic evidence for two independent Ries and Steinheim asteroid impacts in the Middle Miocene. Sci Rep 12: 18603. https://doi.org/10.1038/s41598-022-21409-8 doi: 10.1038/s41598-022-21409-8
    [13] Collins G, Melosh HJ, Marcus R (2005) Earth impact effects program: a web-based computer program for calculating the regional environmental consequences of a meteoroid impact on Earth. Meteorit Planet Sci 40: 817–840. https://doi.org/10.1111/j.1945-5100.2005.tb00157.x doi: 10.1111/j.1945-5100.2005.tb00157.x
    [14] Schmieder M, Sach VJ, Buchner E (2021) The Chöpfi pinnacles near Winterthur, Switzerland: Long-distance effects of the Ries impact-earthquake? Int J Earth Sci 111: 145–147. https://doi.org/10.1007/s00531-021-02082-0 doi: 10.1007/s00531-021-02082-0
    [15] Buchner E, Sach VJ, Schmieder M (2021) Sand spikes pinpoint powerful palaeoseismicity. Nat Commun 12: 6731. https://doi.org/10.1038/s41467-021-27061-6 doi: 10.1038/s41467-021-27061-6
    [16] Maurer H, Buchner E (2007) Rekonstruktion fluvialer Systeme der Oberen Süßwassermolasse im Nordalpinen Vorlandbecken SW-Deutschlands. German J Geosci (ZdGG) 158: 249–270. https://doi.org/10.1127/1860-1804/2007/0158-0249 doi: 10.1127/1860-1804/2007/0158-0249
    [17] Heider J, Wegele A, Amstutz GC (1976) Beobachtungen über Sandrosen und Zapfensande aus der Süßwassermolasse Südwürttembergs. Der Aufschluß 27: 297–307.
    [18] Sanborn WB (1976) Oddities of the Mineral World. Van Nostrand Reinhold Company, NY, USA. 142. Available from: http://allanmccollum.net/amcimages/sanborn.html, (last accessed February 19, 2025).
    [19] Akçiz SO, Grant Ludwig L, Arrowsmith JR, et al. (2010) Century-long average time intervals between earthquake ruptures of the San Andreas fault in the Carrizo Plain, California. Geology 38: 787–790. https://doi.org/10.1130/G30995.1 doi: 10.1130/G30995.1
    [20] McBride EF, Picard MD, Folk RL (1994) Orientated Concretions, Ionian Coast, Italy: Evidence of Groundwater flow direction. J Sediment Res A64: 535–540. https://doi.org/10.1306/D4267DFC-2B26-11D7-8648000102C1865D doi: 10.1306/D4267DFC-2B26-11D7-8648000102C1865D
    [21] McCullough LN, Ritter JB, Zaleha MJ, et al. (2003) Habit, formation, and implications of elongeate, calcite concretions, Victoria, Australia. Department of Geology, Wittenberg University, Ohio, USA. Published Senior Honors Thesis. 25.
    [22] Grant JA, Wilson SA (2018) Possible Geomorphic and Crater Density Evidence for Late Aqueous Activity in Gale Crater. LPI Contrib. 49th Annual Lunar and Planetary Science Conference.
    [23] Metz J, Grotzinger J, Okubo C, et al. (2010) Thin‐skinned deformation of sedimentary rocks in Valles Marineris, Mars. J Geophys Res Planets 115: E11004. https://doi.org/10.1029/2010JE003593 doi: 10.1029/2010JE003593
    [24] NASA Mars Science Laboratory, Curiosity Rover, 2024. Available from: https://mars.nasa.gov/msl/mission/science/.
    [25] Wray JJ (2013) Gale Crater: The Mars Science Laboratory/Curiosity rover landing site. Int J Astrobiol 12: 25–38. https://doi.org/10.1017/S1473550412000328 doi: 10.1017/S1473550412000328
    [26] Grotzinger JP, Sumner DY, Kah LC, et al. (2014) A habitable fluvio-lacustrine environment at Yellowknife Bay, Gale Crater, Mars. Science 343: 6169. https://doi.org/10.1126/science.1242777 doi: 10.1126/science.1242777
    [27] Grotzinger JP, Gupta S, Malin MC, et al. (2015) Deposition, exhumation, and paleoclimate of an ancient lake deposit, Gale Crater, Mars. Science 350: 6257. https://doi.org/10.1126/science.aac7575 doi: 10.1126/science.aac7575
    [28] Buz J, Ehlmann BL, Pan L, et al. (2017) Mineralogy and stratigraphy of the Gale crater rim, wall, and floor units. J Geophys Res Planets 122: 1090–1118. https://doi.org/10.1002/2016JE005163 doi: 10.1002/2016JE005163
    [29] Schwenzer SP, Abramov O, Allen CC, et al. (2012) Gale Crater: Formation and post-impact hydrous environments. Planet Space Sci 70: 84–95. https://doi.org/10.1016/j.pss.2012.05.014 doi: 10.1016/j.pss.2012.05.014
    [30] Montenat C, Barrier P, Ott d'Estevou P, et al. (2007) Seismites: An attempt at critical analysis and classification. Sediment Geol 196: 5–30. https://doi.org/10.1016/j.sedgeo.2006.08.004 doi: 10.1016/j.sedgeo.2006.08.004
    [31] Hargitai H, Levi T (2015) Clastic dikes, Encyclopedia of Planetary Landforms, Hargitai H and Kereszturi A, Eds., Encyclopedia of Planetary Landforms, Springer, NY, USA.
    [32] Sleep NH, Olds EP (2018) Remote faulting triggered by strong seismic waves from the Cretaceous-Paleogene asteroid impact. Seismol Res Lett 89: 570–576. https://doi.org/10.1785/0220170223 doi: 10.1785/0220170223
    [33] DePalma RA, Smit J, Burnham DA, et al. (2019) A seismically induced onshore surge deposit at the K-Pg. boundary, North Dakota. PNAS 116: 8190–8199. https://doi.org/10.1073/pnas.1817407116 doi: 10.1073/pnas.1817407116
    [34] Vaniman DT, Bish DL, Ming DW, et al. (2014) Mineralogy of a mudstone at Yellowknife Bay, Gale Crater, Mars. Science 343: 6169. https://doi.org/10.1126/science.1243480 doi: 10.1126/science.1243480
    [35] Ehlmann BL, Buz J (2015) Mineralogy and fluvial history of the watersheds of Gale, Knobel, and Sharp craters: A regional context for MSL Curiosity's exploration. Geophys Res Lett 42: 264–273. https://doi.org/10.1002/2014GL062553 doi: 10.1002/2014GL062553
    [36] Carter J, Viviano-Beck C, Loizeau D, et al. (2015) Orbital detection and implications of akaganéite on Mars. Icarus 253: 296–310. https://doi.org/10.1016/j.icarus.2015.01.020 doi: 10.1016/j.icarus.2015.01.020
    [37] Tohver E, Schmieder M, Lana C, et al. (2018) End-Permian impactogenic earthquake and tsunami deposits in the intracratonic Paraná Basin of Brazil. GSA Bull 130: 1099–1120. https://doi.org/10.1130/B31626.1 doi: 10.1130/B31626.1
    [38] Weatherley DK, Henley RW (2013) Flash vaporization during earthquakes evidenced by gold deposits. Nature Geosci 6: 294–298. https://doi.org/10.1038/ngeo1759 doi: 10.1038/ngeo1759
    [39] Simms JM (2003) Uniquely extensive seismite from the latest Triassic of the United Kingdom: evidence for bolide impact? Geology 31: 557–560. https://doi.org/10.1130/0091-7613(2003)031<0557:UESFTL>2.0.CO;2 doi: 10.1130/0091-7613(2003)031<0557:UESFTL>2.0.CO;2
    [40] Banham SG, Gupta S, Rubin DM, et al. (2018) Ancient Martian aeolian processes and palaeomorphology reconstructed from the Stimson formation on the lower slope of Aeolis Mons, Gale crater, Mars. Sedimentology 65: 993–1042. https://doi.org/10.1111/sed.12469 doi: 10.1111/sed.12469
    [41] Bohacs KM, Carrol AR, Neal JE (2003) Lessons from large lake systems—Thresholds, nonlinearity, and strange attractors. Special Papers-Geological Society of America, 75–90. https://doi.org/10.1130/0-8137-2370-1.75 doi: 10.1130/0-8137-2370-1.75
    [42] Brož P, Oehler D, Mazzini A, et al. (2023) An overview of sedimentary volcanism on Mars. Earth Surf Dynam 11: 633–661. https://doi.org/10.5194/egusphere-2022-1458 doi: 10.5194/egusphere-2022-1458
    [43] Sturm S, Wulf G, Jung D, et al. (2013) The Ries impact, a double-layer rampart crater on Earth. Geology 41: 531–534. https://doi.org/10.1130/G33934.1 doi: 10.1130/G33934.1
    [44] Wilson SA, Morgan AM, Howard AD, et al. (2021) The global distribution of craters with alluvial fans and deltas on Mars. Geophys Res Lett 48: e2020GL091653. https://doi.org/10.1029/2020GL091653 doi: 10.1029/2020GL091653
    [45] Grotzinger JP, Crisp J, Vasavada AR, et al. (2012) Mars Science Laboratory mission and science investigation. Space Sci Rev 170: 5–56. https://doi.org/10.1007/s11214-012-9892-2 doi: 10.1007/s11214-012-9892-2
  • This article has been cited by:

    1. Ji-Cai Liu, On two supercongruences for sums of Apéry-like numbers, 2021, 115, 1578-7303, 10.1007/s13398-021-01092-6
    2. Rong-Hua Wang, Michael X.X. Zhong, q-Rational reduction and q-analogues of series for π, 2023, 116, 07477171, 58, 10.1016/j.jsc.2022.08.020
    3. Qing-hu Hou, Guo-jie Li, Gosper summability of rational multiples of hypergeometric terms, 2021, 27, 1023-6198, 1723, 10.1080/10236198.2021.2007903
    4. Qing-Hu Hou, Ke Liu, Congruences and telescopings of P-recursive sequences, 2021, 27, 1023-6198, 686, 10.1080/10236198.2021.1934462
    5. Ji-Cai Liu, On two congruences involving Franel numbers, 2020, 114, 1578-7303, 10.1007/s13398-020-00935-y
    6. Liuquan Wang, Yifan Yang, Ramanujan-type 1/\pi -series from bimodular forms, 2022, 59, 1382-4090, 831, 10.1007/s11139-021-00532-6
    7. Zhi-Wei Sun, On Motzkin numbers and central trinomial coefficients, 2022, 136, 01968858, 102319, 10.1016/j.aam.2021.102319
    8. Ji-Cai Liu, Ramanujan-Type Supercongruences Involving Almkvist–Zudilin Numbers, 2022, 77, 1422-6383, 10.1007/s00025-022-01607-6
    9. Qing-Hu Hou, Zhi-Wei Sun, q-Analogues of Some Series for Powers of \pi , 2021, 25, 0218-0006, 167, 10.1007/s00026-021-00522-x
    10. Rong-Hua Wang, Rational Reductions for Holonomic Sequences, 2024, 1009-6124, 10.1007/s11424-024-4034-y
    11. Chunli Li, Wenchang Chu, Infinite series about harmonic numbers inspired by Ramanujan–like formulae, 2023, 31, 2688-1594, 4611, 10.3934/era.2023236
    12. Zhi-Wei Sun, 2025, Chapter 21, 978-3-031-65063-5, 413, 10.1007/978-3-031-65064-2_21
    13. Sun Zhi-Wei, Infinite series involving binomial coefficients and harmonic numbers, 2024, 54, 1674-7216, 765, 10.1360/SSM-2024-0007
  • Reader Comments
  • © 2025 the Author(s), licensee AIMS Press. This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0)
通讯作者: 陈斌, bchen63@163.com
  • 1. 

    沈阳化工大学材料科学与工程学院 沈阳 110142

  1. 本站搜索
  2. 百度学术搜索
  3. 万方数据库搜索
  4. CNKI搜索

Metrics

Article views(661) PDF downloads(64) Cited by(1)

Figures and Tables

Figures(6)

Other Articles By Authors

/

DownLoad:  Full-Size Img  PowerPoint
Return
Return

Catalog