Processing math: 68%
Research article Special Issues

A flexible model for bounded data with bathtub shaped hazard rate function and applications

  • The unit new X-Lindley distribution, which is a novel one-parameter distribution on the unit interval, is presented in this study. It was developed by altering the new X-Lindley distribution using the exponential transformation. This new one-parameter distribution's fundamental features, including moments, incomplete moments, Lorenz and Bonferroni curves, Gini index, mode, extropy, Havrda and Charvat entropy, Rényi entropy, and Tsallis entropy, are explored. Additionally, it has bathtub-shaped hazard rate functions and monotonically increasing hazard rate functions with a single parameter. The new distribution is therefore sufficiently rich to model real data. Also, different estimation methods, such as maximum likelihood, least-squares, and weighted least-squares, are used to estimate the parameters of this model, and using a simulation research, their respective performances are evaluated. Finally, two real-life datasets are used to demonstrate the suggested model's competency.

    Citation: M. R. Irshad, S. Aswathy, R. Maya, Amer I. Al-Omari, Ghadah Alomani. A flexible model for bounded data with bathtub shaped hazard rate function and applications[J]. AIMS Mathematics, 2024, 9(9): 24810-24831. doi: 10.3934/math.20241208

    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 unit new X-Lindley distribution, which is a novel one-parameter distribution on the unit interval, is presented in this study. It was developed by altering the new X-Lindley distribution using the exponential transformation. This new one-parameter distribution's fundamental features, including moments, incomplete moments, Lorenz and Bonferroni curves, Gini index, mode, extropy, Havrda and Charvat entropy, Rényi entropy, and Tsallis entropy, are explored. Additionally, it has bathtub-shaped hazard rate functions and monotonically increasing hazard rate functions with a single parameter. The new distribution is therefore sufficiently rich to model real data. Also, different estimation methods, such as maximum likelihood, least-squares, and weighted least-squares, are used to estimate the parameters of this model, and using a simulation research, their respective performances are evaluated. Finally, two real-life datasets are used to demonstrate the suggested model's competency.



    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

    limnn|Tn(b,c)|=b24c.

    In 2011, the author posed over 60 conjectural series for 1/π of the following new types with a,b,c,d,m integers and mbcd(b24c) nonzero (cf. Sun [34,40]).

    Type Ⅰ. k=0a+dkmk(2kk)2Tk(b,c).

    Type Ⅱ. k=0a+dkmk(2kk)(3kk)Tk(b,c).

    Type Ⅲ. k=0a+dkmk(4k2k)(2kk)Tk(b,c).

    Type Ⅳ. k=0a+dkmk(2kk)2T2k(b,c).

    Type Ⅴ. k=0a+dkmk(2kk)(3kk)T3k(b,c).

    Type Ⅵ. k=0a+dkmkTk(b,c)3,

    Type Ⅶ. k=0a+dkmk(2kk)Tk(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

    k=03990k+1147(288)3kTk(62,952)3=43295π(942+19514)

    as well as its p-adic analogue

    p1k=03990k+1147(288)3kTk(62,952)3p19(4230(2p)+17563(14p)) (mod p2),

    where p is any prime greater than 3.

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

    k=0(4k1)(2kk)4(2k1)4256k=8π2.

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

    nk=0(4k1)(2kk)4(2k1)4256k=(8n2+4n+1)(2nn)4256n for all nN.

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

    Theorem 1.1. We have the following identities:

    k=0k(4k1)(2kk)3(2k1)2(64)k=1π, (1.1)
    k=0(4k1)(2kk)3(2k1)3(64)k=2π, (1.2)
    k=0(12k21)(2kk)3(2k1)2256k=2π, (1.3)
    k=0k(6k1)(2kk)3(2k1)3256k=12π, (1.4)
    k=0(28k24k1)(2kk)3(2k1)2(512)k=32π, (1.5)
    k=0(30k2+3k2)(2kk)3(2k1)3(512)k=2728π, (1.6)
    k=0(28k24k1)(2kk)3(2k1)24096k=3π, (1.7)
    k=0(42k23k1)(2kk)3(2k1)34096k=278π, (1.8)
    k=0(34k23k1)(2kk)2(3kk)(2k1)(3k1)(192)k=103π, (1.9)
    k=0(64k211k7)(2kk)2(3kk)(k+1)(2k1)(3k1)(192)k=12539π, (1.10)
    k=0(14k2+k1)(2kk)2(3kk)(2k1)(3k1)216k=3π, (1.11)
    k=0(90k2+7k+1)(2kk)2(3kk)(k+1)(2k1)(3k1)216k=932π, (1.12)
    k=0(34k23k1)(2kk)2(3kk)(2k1)(3k1)(12)3k=23π, (1.13)
    k=0(17k+5)(2kk)2(3kk)(k+1)(2k1)(3k1)(12)3k=93π, (1.14)
    k=0(111k27k4)(2kk)2(3kk)(2k1)(3k1)1458k=454π, (1.15)
    k=0(1524k2+899k+263)(2kk)2(3kk)(k+1)(2k1)(3k1)1458k=33754π, (1.16)
    k=0(522k255k13)(2kk)2(3kk)(2k1)(3k1)(8640)k=54155π, (1.17)
    k=0(1836k2+2725k+541)(2kk)2(3kk)(k+1)(2k1)(3k1)(8640)k=2187155π, (1.18)
    k=0(529k245k16)(2kk)2(3kk)(2k1)(3k1)153k=5532π, (1.19)
    k=0(77571k2+68545k+16366)(2kk)2(3kk)(k+1)(2k1)(3k1)153k=5989532π, (1.20)
    k=0(574k273k11)(2kk)2(3kk)(2k1)(3k1)(48)3k=203π, (1.21)
    k=0(8118k2+9443k+1241)(2kk)2(3kk)(k+1)(2k1)(3k1)(48)3k=22503π, (1.22)
    k=0(978k2131k17)(2kk)2(3kk)(2k1)(3k1)(326592)k=990749π, (1.23)
    k=0(592212k2+671387k2+77219)(2kk)2(3kk)(k+1)(2k1)(3k1)(326592)k=4492125749π, (1.24)
    k=0(116234k217695k1461)(2kk)2(3kk)(2k1)(3k1)(300)3k=26503π, (1.25)
    k=0(223664832k2+242140765k+18468097)(2kk)2(3kk)(k+1)(2k1)(3k1)(300)3k=334973253π, (1.26)
    k=0(122k2+3k5)(2kk)2(4k2k)(2k1)(4k1)648k=212π, (1.27)
    k=0(1903k2+114k+41)(2kk)2(4k2k)(k+1)(2k1)(4k1)648k=3432π, (1.28)
    k=0(40k22k1)(2kk)2(4k2k)(2k1)(4k1)(1024)k=4π, (1.29)
    k=0(8k22k1)(2kk)2(4k2k)(k+1)(2k1)(4k1)(1024)k=165π, (1.30)
    k=0(176k26k5)(2kk)2(4k2k)(2k1)(4k1)482k=83π, (1.31)
    k=0(208k2+66k+23)(2kk)2(4k2k)(k+1)(2k1)(4k1)482k=1283π, (1.32)
    k=0(6722k2411k152)(2kk)2(4k2k)(2k1)(4k1)(632)k=1957π, (1.33)
    k=0(281591k2757041k231992)(2kk)2(4k2k)(k+1)(2k1)(4k1)(632)k=2746257π, (1.34)
    k=0(560k242k11)(2kk)2(4k2k)(2k1)(4k1)124k=242π, (1.35)
    k=0(112k2+114k+23)(2kk)2(4k2k)(k+1)(2k1)(4k1)124k=25625π, (1.36)
    k=0(248k218k5)(2kk)2(4k2k)(2k1)(4k1)(3×212)k=283π, (1.37)
    k=0(680k2+1482k+337)(2kk)2(4k2k)(k+1)(2k1)(4k1)(3×212)k=548839π, (1.38)
    k=0(1144k2102k19)(2kk)2(4k2k)(2k1)(4k1)(21034)k=60π, (1.39)
    k=0(3224k2+4026k+637)(2kk)2(4k2k)(k+1)(2k1)(4k1)(21034)k=2000π, (1.40)
    k=0(7408k2754k103)(2kk)2(4k2k)(2k1)(4k1)284k=56033π, (1.41)
    k=0(3641424k2+4114526k+493937)(2kk)2(4k2k)(k+1)(2k1)(4k1)284k=8960003π, (1.42)
    k=0(4744k2534k55)(2kk)2(4k2k)(2k1)(4k1)(214345)k=1932525π, (1.43)
    k=0(18446264k2+20356230k+1901071)(2kk)2(4k2k)(k+1)(2k1)(4k1)(214345)k=66772496525π, (1.44)
    k=0(413512k250826k3877)(2kk)2(4k2k)(2k1)(4k1)(210214)k=12180π, (1.45)
    k=0(1424799848k2+1533506502k+108685699)(2kk)2(4k2k)(k+1)(2k1)(4k1)(210214)k=341446000π, (1.46)
    k=0(71312k27746k887)(2kk)2(4k2k)(2k1)(4k1)15842k=84011π, (1.47)
    k=0(50678512k2+56405238k+5793581)(2kk)2(4k2k)(k+1)(2k1)(4k1)15842k=548800011π, (1.48)
    k=0(7329808k2969294k54073)(2kk)2(4k2k)(2k1)(4k1)3964k=1201202π, (1.49)
    k=0(2140459883152k2+2259867244398k+119407598201)(2kk)2(4k2k)(k+1)(2k1)(4k1)3964k=44×182032π, (1.50)
    k=0(164k2k3)(2kk)(3kk)(6k3k)(2k1)(6k1)203k=752π, (1.51)
    k=0(2696k2+206k+93)(2kk)(3kk)(6k3k)(k+1)(2k1)(6k1)203k=6865π, (1.52)
    k=0(220k28k3)(2kk)(3kk)(6k3k)(2k1)(6k1)(215)k=72π, (1.53)
    k=0(836k21048k309)(2kk)(3kk)(6k3k)(k+1)(2k1)(6k1)(215)k=6862π, (1.54)
    k=0(504k211k8)(2kk)(3kk)(6k3k)(2k1)(6k1)(15)3k=915π, (1.55)
    k=0(189k211k8)(2kk)(3kk)(6k3k)(k+1)(2k1)(6k1)(15)3k=2431535π, (1.56)
    k=0(516k219k7)(2kk)(3kk)(6k3k)(2k1)(6k1)(2×303)k=11152π, (1.57)
    k=0(3237k2+1922k+491)(2kk)(3kk)(6k3k)(k+1)(2k1)(6k1)(2×303)k=39931510π, (1.58)
    k=0(684k240k7)(2kk)(3kk)(6k3k)(2k1)(6k1)(96)3k=96π, (1.59)
    k=0(2052k2+2536k+379)(2kk)(3kk)(6k3k)(k+1)(2k1)(6k1)(96)3k=4866π, (1.60)
    k=0(2556k2131k29)(2kk)(3kk)(6k3k)(2k1)(6k1)663k=63334π, (1.61)
    k=0(203985k2+212248k+38083)(2kk)(3kk)(6k3k)(k+1)(2k1)(6k1)663k=83349334π, (1.62)
    k=0(5812k2408k49)(2kk)(3kk)(6k3k)(2k1)(6k1)(3×1603)k=253309π, (1.63)
    k=0(3471628k2+3900088k+418289)(2kk)(3kk)(6k3k)(k+1)(2k1)(6k1)(3×1603)k=3238855430135π, (1.64)
    k=0(35604k22936k233)(2kk)(3kk)(6k3k)(2k1)(6k1)(960)3k=18915π, (1.65)
    k=0(13983084k2+15093304k+1109737)(2kk)(3kk)(6k3k)(k+1)(2k1)(6k1)(960)3k=4500846155π, (1.66)
    k=0(157752k211243k1304)(2kk)(3kk)(6k3k)(2k1)(6k1)2553k=5132552π, (1.67)
    k=0(28240947k2+31448587k+3267736)(2kk)(3kk)(6k3k)(k+1)(2k1)(6k1)2553k=4500189925570π, (1.68)
    k=0(2187684k2200056k11293)(2kk)(3kk)(6k3k)(2k1)(6k1)(5280)3k=1953330π, (1.69)
    k=0(101740699836k2+107483900696k+5743181813)(2kk)(3kk)(6k3k)(k+1)(2k1)(6k1)(5280)3k=49661001183305π, (1.70)
    k=0(16444841148k21709536232k53241371)(2kk)(3kk)(6k3k)(2k1)(6k1)(640320)3k=167220910005π, (1.71)

    and

    k=0P(k)(2kk)(3kk)(6k3k)(k+1)(2k1)(6k1)(640320)3k=18×5574033100055π, (1.72)

    where

    P(k):=637379600041024803108k2+657229991696087780968k+19850391655004126179.

    Recall that the Catalan numbers are given by

    Cn:=(2nn)n+1=(2nn)(2nn+1)  (nN).

    For kN it is easy to see that

    (2kk)2k1={1if k=0,2Ck1if k>0.

    Thus, for any a,b,c,mZ with |m|64, we have

    k=0(ak2+bk+c)(2kk)3(2k1)3mk=c+k=1(ak2+bk+c)(2Ck1)3mk=c+8mk=0a(k+1)2+b(k+1)+cmkC3k.

    For example, (1.2) has the equivalent form

    k=04k+3(64)kC3k=816π.(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

    (p+1)/2k=0(4k1)(2kk)3(2k1)3(64)kp(1p)+p3(Ep32) (mod p4)

    (where E0,E1, are the Euler numbers), and we note that this is also equivalent to the congruence

    (p1)/2k=04k+3(64)kC3k8(1p(1p)p3(Ep32)) (mod p4).

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

    (p+1)/2k=0(3k1)(2kk)3(2k1)216kp+2p3(1p)(Ep33) (mod p4)

    and

    p1k=0(3k1)(2kk)3(2k1)216kp2p3 (mod p4).

    (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

    k=0(56k2+118k+61)(2kk)3(k+1)24096k=192π

    and

    k=0(420k2+992k+551)(2kk)3(k+1)2(2k1)4096k=1728π.

    Now we state our second theorem.

    Theorem 1.2. We have the identities

    k=128k2+31k+8(2k+1)2k3(2kk)3=π282, (1.73)
    k=142k2+39k+8(2k+1)3k3(2kk)3=9π2882, (1.74)
    k=1(8k2+5k+1)(8)k(2k+1)2k3(2kk)3=46G, (1.75)
    k=1(30k2+33k+7)(8)k(2k+1)3k3(2kk)3=54G52, (1.76)
    k=1(3k+1)16k(2k+1)2k3(2kk)3=π282, (1.77)
    k=1(4k+1)(64)k(2k+1)2k2(2kk)3=48G, (1.78)
    k=1(4k+1)(64)k(2k+1)3k3(2kk)3=16G16, (1.79)
    k=1(2k211k3)8k(2k+1)(3k+1)k3(2kk)2(3kk)=485π22, (1.80)
    k=2(178k2103k39)8k(k1)(2k+1)(3k+1)k3(2kk)2(3kk)=1125π21109636, (1.81)
    k=1(5k+1)(27)k(2k+1)(3k+1)k2(2kk)2(3kk)=69K, (1.82)
    k=2(45k2+5k2)(27)k1(k1)(2k+1)(3k+1)k3(2kk)2(3kk)=3748K16, (1.83)
    k=1(98k221k8)81k(2k+1)(4k+1)k3(2kk)2(4k2k)=21620π2, (1.84)
    k=2(1967k2183k104)81k(k1)(2k+1)(4k+1)k3(2kk)2(4k2k)=20000π2190269120, (1.85)
    k=1(46k2+3k1)(144)k(2k+1)(4k+1)k3(2kk)2(4k2k)=722252K, (1.86)
    k=2(343k2+18k16)(144)k(k1)(2k+1)(4k+1)k3(2kk)2(4k2k)=9375K704810, (1.87)

    where

    G:=k=0(1)k(2k+1)2  and  K:=k=0(k3)k2.

    For k=j+1Z+, it is easy to see that

    (k1)k(2kk)=2(2j+1)j(2jj).

    Thus, for any a,b,c,mZ with 0<|m|64, we have

    j=1(aj2+bj+c)mj(2j+1)3j3(2jj)3=8mk=2(a(k1)2+b(k1)+c)mk(k1)3k3(2kk)3.

    For example, (1.77) has the following equivalent form

    k=2(2k1)(3k2)16k(k1)3k3(2kk)3=π28.(1.77)

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

    Sn:=nk=0(nk)2TkTnk  (n=0,1,2,).

    The values of Sn (n=0,,10) are

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

    respectively. We may extend the numbers Sn (nN) further. For b,cZ, we define

    Sn(b,c):=nk=0(nk)2Tk(b,c)Tnk(b,c)  (n=0,1,2,).

    Note that Sn(1,1)=Sn and Sn(2,1)=Dn for all nN.

    Now we state our third theorem.

    Theorem 1.3. We have

    k=07k+324kSk(1,6)=152π, (1.88)
    k=012k+5(28)kSk(1,7)=67π, (1.89)
    k=084k+2980kSk(1,20)=2415π, (1.90)
    k=03k+1(100)kSk(1,25)=258π, (1.91)
    k=0228k+67224kSk(1,56)=807π, (1.92)
    k=0399k+101(676)kSk(1,169)=25358π, (1.93)
    k=02604k+5632600kSk(1,650)=850393π, (1.94)
    k=039468k+7817(6076)kSk(1,1519)=441031π, (1.95)
    k=041667k+78799800kSk(1,2450)=4042564π, (1.96)
    k=074613k+10711(5302)kSk(1,2652)=161517548π. (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 π involving generalized central trinomial coefficients. In particular, we will present in Section 3 four conjectural series for 1/π of the following new type:

    Type Ⅷ. k=0a+dkmkTk(b,c)Tk(b,c)2,

    where a,b,b,c,c,d,m are integers with mbbccd(b24c)(b24c)(b2cb2c)0.

    Unlike Ramanujan-type series given by others, all our series for 1/π 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/π (cf. Sun [35]), we formulate the following general characterization of rational Ramanujan-type series for 1/π via congruences.

    Conjecture 1.1 (General Criterion for Rational Ramanujan-type Series for 1/π). Suppose that the series k=0bk+cmkak converges, where (ak)k0 is an integer sequence and b,c,m are integers with bcm0. Suppose also that there are no a,xZ such that an=a(2nn)nk=0(2kk)2(2(nk)nk)xnk for all nN. Let r{1,2,3} and let d1,,drZ+ with di/dj irrational for all distinct i,j{1,,r}. Then

    k=0bk+cmkak=ri=1λidiπ (1.98)

    for some nonzero rational numbers λ1,,λr if and only if there are positive integers dj (r<j3) and rational numbers c1,c2,c3 with ri=1ci0, such that for any prime p>3 with pmri=1di and c1,c2,c3Zp we have

    p1k=0bk+cmkakp(ri=1ci(εidip)+r<j3cj(djp)) (mod p2), (1.99)

    where εi{±1}, εi=1 if di is not an integer square, and c2=c3=0 if r=1 and ε1=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 (an)n0 be an integer sequence with no a,xZ such that an=a(2nn)nk=0(2kk)2(2(nk)nk)xnk for all nN, and let b,c,m,d1,d2,d3Z with bcm0. Assume that limn+n|an|=r<|m|, and πk=0bk+cmkak is an algebraic number. Suppose that c1,c2,c3Q with c1+c2+c3=a0c , and

    p1k=0bk+cmkakp(c1(d1p)+c2(d2p)+c3(d3p)) (mod p2) (1.100)

    for all primes p>3 with pd1d2d3m and c1,c2,c3Zp. Then, for any prime p>3 with pm, c1,c2,c3Zp and (d1p)=(d2p)=(d3p)=δ{±1}, we have

    1(pn)2(pn1k=0bk+cmkakpδn1k=0bk+cmkak)Zp  for all nZ+.

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

    Conjecture 1.3 (Chen Wang and Z.-W. Sun). Let (ak)k0 be an integer sequence with a0=1. Let b,c,m,d1,d2,d3Z with bm0, and let c1,c2,c3 be rational numbers. If πk=0bk+cmkak is an algebraic number, and the congruence (1.100) holds for all primes p>3 with pd1d2d3m and c1,c2,c3Zp, then we must have c1+c2+c3=c.

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

    p1k=0(8k+5)T2k3p(3p) (mod p2)

    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 k=0(8k+5)T2k 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/π in Sections 3-9 based on the author's previous PhilosophyaboutSeriesfor 1/π} stated in [35], the PSLQ algorithm to discover integer relations (cf. [13]), and the following DualityPrinciple based on the author's experience and intuition.

    Conjecture 1.4 (Duality Principle). Let (ak)k0 be an integer sequence such that

    ak(dp)Dkap1k (mod p) (1.101)

    for any prime p6dD and k{0,,p1}, where d and D are fixed nonzero integers. If a0,a1, are not all zero and m is a nonzero integer such that

    k=0bk+cmkak=λ1d1+λ2d2+λ3d3π

    for some b,d1,d2,d3Z+, cZ and λ1,λ2,λ3Q, then m divides D, and

    p1k=0akmk(dp)p1k=0ak(D/m)k (mod p2) (1.102)

    for any prime p>3 with pdD.

    Remark 1.3 (ⅰ) For any prime p>3 with pdDm, the congruence (1.102) holds modulo p by (1.101) and Fermat's little theorem. We call p1k=0ak/(Dm)k the dual of the sum p1k=0ak/mk.

    (ⅱ) For any b,cZ and odd prime pb24c, it is known (see, e.g., [39,Lemma 2.2]) that

    Tk(b,c)(b24cp)(b24c)kTp1k(b,c) (mod p) (1.103)

    for all k=0,1,,p1.

    For a series k=0ak with a0,a1, real numbers, if limk+ak+1/ak=r(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 π involving the central trinomial coefficients.

    Lemma 2.1. Let m0 and n0 be integers. Then

    nk=0((64m)k332k216k+8)(2kk)3(2k1)2mk=8(2n+1)mn(2nn)3, (2.1)
    nk=0((64m)k396k2+48k8)(2kk)3(2k1)3mk=8mn(2nn)3, (2.2)
    nk=0((108m)k354k212k+6)(2kk)2(3kk)(2k1)(3k1)mk=6(3n+1)mn(2nn)2(3nn), (2.3)
    nk=0((108m)k3(54+m)k212k+6)(2kk)2(3kk)(k+1)(2k1)(3k1)mk=6(3n+1)(n+1)mn(2nn)2(3nn), (2.4)
    nk=0((256m)k3128k216k+8)(2kk)2(4k2k)(2k1)(4k1)mk=8(4n+1)mn(2nn)2(4n2n), (2.5)
    nk=0((256m)k3(128+m)k216k+8)(2kk)2(4k2k)(k+1)(2k1)(4k1)mk=8(4n+1)(n+1)mn(2nn)2(4n2n), (2.6)
    nk=0((1728m)k3864k248k+24)(2kk)(3kk)(6k3k)(2k1)(6k1)mk=24(6n+1)mn(2nn)(3nn)(6n3n), (2.7)
    nk=0((1728m)k3(864+m)k248k+24)(2kk)(3kk)(6k3k)(k+1)(2k1)(6k1)mk=24(6n+1)(n+1)mn(2nn)(3nn)(6n3n). (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!2πn(n/e)n as n+, we have

    (2nn)4nnπ,  (2nn)(3nn)327n2nπ, (2.9)
    (2nn)(4n2n)64n2nπ,  (3nn)(6nn)432n2nπ. (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

    k=0(16k34k22k+1)(2kk)3(2k1)2(64)k=limn+2n+1(64)n(2nn)3=0.

    Note that

    16k34k22k+1=(4k+1)(2k1)2+2k(4k1)

    and recall Bauer's series

    k=0(4k+1)(2kk)3(64)k=2π.

    So, we get

    k=0k(4k1)(2kk)3(2k1)2(64)k=12k=0(4k+1)(2kk)3(64)k=1π.

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

    nk=0(4k1)(4k22k+1)(2kk)3(2k1)3(64)k=(2nn)3(64)n

    and hence

    k=0(2k(2k1)(4k1)+4k1)(2kk)3(2k1)3(64)k=limn+(2nn)3(64)n=0.

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

    In view of (2.7) with m=6403203, we have

    nk=0(10939058860032072k336k22k+1)(2kk)(3kk)(6k3k)(2k1)(6k1)(640320)3k=6n+1(640320)3n(2nn)(3nn)(6n3n).

    and hence

    k=0(10939058860032072k336k22k+1)(2kk)(3kk)(6k3k)(2k1)(6k1)(640320)3k=0.

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

    k=0545140134k+13591409(640320)3k(2kk)(3kk)(6k3k)=3×5336022π10005,

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

    10939058860032072k336k22k+1=1672209(2k1)(6k1)(545140134k+13591409)+426880(16444841148k21709536232k53241371)

    and hence

    k=0(16444841148k21709536232k53241371)(2kk)(3kk)(6k3k)(2k1)(6k1)(640320)3k=1672209426880×3×5336022π10005=167220910005π.

    This proves (1.71).

    By (2.8) with m=6403203, we have

    nk=0(10939058860032072k3+10939058860031964k22k+1)(2kk)(3kk)(6k3k)(k+1)(2k1)(6k1)(640320)3k=6n+1(n+1)(640320)3n(2nn)(3nn)(6n3n)

    and hence

    k=0(10939058860032072k3+10939058860031964k22k+1)(2kk)(3kk)(6k3k)(k+1)(2k1)(6k1)(640320)3k=0.

    Note that

    2802461(10939058860032072k3+10939058860031964k22k+1)=1864188626454(k+1)(16444841148k21709536232k53241371)+5P(k).

    Therefore, with the help of (1.71) we get

    k=0P(k)(2kk)(3kk)(6k3k)(k+1)(2k1)(6k1)(640320)3k=18641886264545×(1672209)10005π=18×5574033100055π.

    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

    nk=1mk((m64)k332k2+16k+8)(2k+1)2k3(2kk)3=mn+1(2n+1)2(2nn)3m, (2.11)
    nk=1mk((m64)k396k248k8)(2k+1)3k3(2kk)3=mn+1(2n+1)3(2nn)3m, (2.12)
    nk=1mk((m108)k354k2+12k+6)(2k+1)(3k+1)k3(2kk)2(3kk)=mn+1(2n+1)(3n+1)(2nn)2(3nn)m, (2.13)
    1<knmk((m108)k3(54+m)k2+12k+6)(k1)(2k+1)(3k+1)k3(2kk)2(3kk)=mn+1n(2n+1)(3n+1)(2nn)2(3nn)m2144, (2.14)
    nk=1mk((m256)k3128k2+16k+8)(2k+1)(4k+1)k3(2kk)2(4k2k)=mn+1(2n+1)(4n+1)(2nn)2(4n2n)m, (2.15)
    1<knmk((m256)k3(128+m)k2+16k+8)(k1)(2k+1)(4k+1)k3(2kk)2(4k2k)=mn+1n(2n+1)(4n+1)(2nn)2(4n2n)m2360. (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:

    k=121k8k3(2kk)3=π26,  k=1(4k1)(64)kk3(2kk)3=16G,k=1(3k1)(8)kk3(2kk)3=2G,  k=1(3k1)16kk3(2kk)3=π22,k=1(15k4)(27)k1k3(2kk)2(3kk)=K,  k=1(5k1)(144)kk3(2kk)2(4k2k)=452K,k=1(11k3)64kk2(2kk)2(3kk)=8π2, k=1(10k3)8kk3(2kk)2(3kk)=π22,  k=1(35k8)81kk3(2kk)2(4k2k)=12π2.

    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

    nk=116k(48k332k2+16k+8)(2k+1)2k3(2kk)3=16n+1(2n+1)2(2nn)316

    for all nZ+, and hence

    k=116k(6k3+4k22k1)(2k+1)2k3(2kk)3=limn+(2×16n(2n+1)2(2nn)3+2)=2.

    Notice that

    2(6k3+4k22k1)=(2k+1)2(3k1)(3k+1).

    So we have

    k=1(3k+1)16k(2k+1)2k3(2kk)3=2×2k=1(3k1)16kk3(2kk)3=4π22

    and hence (1.77) holds.

    By (2.11) with m=64, we have

    nk=1(64)k(128k332k2+16k+8)(2k+1)2k3(2kk)3=(64)n+1(2n+1)2(2nn)3+64

    for all nZ+, and hence

    k=1(64)k(16k3+4k22k1)(2k+1)2k3(2kk)3=8+limn+8(64)n(2n+1)2(2nn)3=8.

    Since 16k3+4k22k1=(4k1)(2k+1)22k(4k+1) and

    k=1(4k1)(64)kk3(2kk)3=16G,

    we see that

    16G2k=1(4k+1)(64)k(2k+1)2k2(2kk)3=8

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

    nk=1(64)k(128k396k248k8)(2k+1)3k3(2kk)3=(64)n+1(2n+1)3(2nn)3+64

    for all nZ+, and hence

    k=1(64)k(16k3+12k2+6k+1)(2k+1)3k3(2kk)3=8+limn+8(64)n(2n+1)3(2nn)3=8.

    Since 16k3+12k2+6k+1=2k(2k+1)(4k+1)+(4k+1), with the aid of (1.78) we obtain

    k=1(4k+1)(64)k(2k+1)3k3(2kk)3=82(48G)=16G16.

    This proves (1.79).

    By (2.13) with m=27, we have

    k=1(45k3+18k24k2)(27)k(2k+1)(3k+1)k3(2kk)2(3kk)=9.

    As

    2(45k3+18k24k2)=(15k4)(2k+1)(3k+1)3k(5k+1)

    and

    k=1(15k4)(27)kk3(2kk)2(3kk)=27K,

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

    3k=2(27)k(45k3+9k24k2)(k1)(2k+1)(3k+1)k3(2kk)2(3kk)=(27)2144

    and hence

    k=2(27)k(45k3+9k24k2)(k1)(2k+1)(3k+1)k3(2kk)2(3kk)=2716.

    As

    45k3+9k24k2=9(k1)k(5k+1)+(45k2+5k2),

    with the aid of (1.82) we get

    k=2(27)k(45k2+5k2)(k1)(2k+1)(3k+1)k3(2kk)2(3kk)=27169(69K6(27)122)=2716(48K37)

    and hence (1.83) follows.

    Other identities in Theorem 1.2 can be proved similarly.

    For integers nk0, we define

    sn,k:=1(nk)ki=0(n2i)(n2(ki))(2ii)(2(ki)ki). (2.17)

    For nN we set

    tn:=0<kn(n1k1)(1)k4nksn+k,k. (2.18)

    Lemma 2.3. For any nN, we have

    nk=0(nk)(1)k4nksn+k,k=fn (2.19)

    and

    (2n+1)tn+1+8ntn=(2n+1)fn+14(n+1)fn, (2.20)

    where fn denotes the Franel number nk=0(nk)3.

    Proof. For n,i,kN with ik, we set

    F(n,i,k)=(nk)(1)k4nk(n+kk)(n+k2i)(2ii)(n+k2(ki))(2(ki)ki).

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

    F(n,i,k):=F(n,i,k)+7n2+21n+168(n+1)2F(n+1,i,k)(n+2)28(n+1)2F(n+2,i,k)

    with 0ik, we find that

    F(n,i,k)=(G1(n,i+1,k)G1(n,i,k))+(G2(n,i,k+1)G2(n,i,k)),

    where

    G1(n,i,k):=i2(k+i1)(1)k+14nkn!2(n+k)!p(n,i,k)(2n+3)(nk+2)!(n+k+22i)!(nk+2i)!(i!(ki+1)!)2

    and

    G2(n,i,k):=2(ki)(1)k4nkn!2(n+k)!q(n,i,k)(2n+3)(nk+2)!(n+k2i+1)!(nk+2i+2)!(i!(ki)!)2,

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

    10n4+(i10k68)n3+(24i2+(32k+31)i+2k267k172)n2+(36i3+(68k124)i2+(39k2+149k+104)i+2k38k2145k192)n+60i3+(114k140)i2+(66k2+160k+92)i+3k319k2102k80

    and

    10(ik)n4+(20i2+(46k+47)i6k247k)n3+(72i3+(60k38)i2+(22k2+145k+90)i+4k311k290k)n2+(72k+156)i3n+(72k260k10)i2n+(18k3+4k2+165k+85)in+(22k35k285k)n+(120k+60)i3+(120k2+68k4)i2+(30k356k2+86k+32)i+26k36k232k

    respectively. Therefore

    n+2k=0ki=0F(n,i,k)=n+2k=0(G1(n,k+1,k)G1(n,0,k))+n+2i=0(G2(n,i,n+3)G2(n,i,i))=n+2k=0(00)+n+2i=0(00)=0,

    and hence

    u(n):=nk=0(nk)(1)k4nksn+k,k

    satisfies the recurrence relation

    8(n+1)2u(n)+(7n2+21n+16)u(n+1)(n+2)2u(n+2)=0.

    As pointed out by J. Franel [14], the Franel numbers satisfy the same recurrence. Note also that u(0)=f0=1 and u(1)=f1=2. So we always have u(n)=fn. 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

    8(n+1)(n+2)(18n3+117n2+249n+172)v(n)+(126n5+1197n4+4452n3+8131n2+7350n+2656)v(n+1)=(n+3)2(18n3+63n2+69n+22)v(n+2).

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

    Lemma 2.4. For any cZ and nN, we have

    Sn(4,c)=n/2k=0(nkk)(2(nk)nk)ck4n2ksn,k. (2.21)

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

    Tk(4,c)Tnk(4,c)=k/2i=0(k2i)(2ii)4k2ici(nk)/2j=0(nk2j)(2jj)4nk2jcj=n/2r=0cr4n2ri,jNi+j=r(k2i)(nk2j)(2ii)(2jj).

    If i,jN and i+j=rn/2, then

    nk=0(nk)2(k2i)(nk2j)=(n2i)(n2j)n2jk=2i(n2ik2i)(n2jnk2j)=(n2i)(n2j)(2n2(i+j)n2(i+j))=(2n2rn)(n2i)(n2j)

    with the aid of the Chu-Vandermonde identity. Therefore

    Sn(4,c)=n/2k=0ck4n2k(2n2kn)(nk)sn,k=n/2k=0ck4n2k(2n2knk)(nkk)sn,k.

    This proves (2.21).

    Lemma 2.5. For kN and lZ+, we have

    sk+l,k(2k+1)4kl(k+ll). (2.22)

    Proof. Let n=k+l. Then

    (nk)sn,ki,jNi+j=k(n2i)(n2j)i,jNi+j=k(2ii)(2jj)s,tNs+t=2k(ns)(nt)i,jNi+j=k4i4j=(2n2k)(k+1)4k

    and

    (2n2k)(nk)=(2n2l)(nl)=l1j=02(j+k)+12j+1(2k+1)0<j<l2(j+k)2j=(2k+1)(k+l1l1).

    Hence

    sk+l,k(k+1)4k(2k+1)lk+l(k+ll)(2k+1)4kl(k+ll).

    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|94, we have

    n=0(an+b)Sn(4,m)mn=1m+16n=0(2a(m+4)n8a+b(m+16))(2nn)fnmn. (2.23)

    Proof. Let NN. In view of (2.21),

    Nn=0Sn(4,m)mn=Nn=01mnn/2k=0(m)k4n2k(2n2knk)(nkk)sn,k
    =Nl=0(2ll)mlNlk=0(lk)(1)k4lksl+k,k=N/2l=0(2ll)mllk=0(lk)(1)k4lksl+k,k+N/2<lN(2ll)mlNlk=0(lk)(1)k4lksl+k,k

    and similarly

    Nn=0nSn(4,m)mn=Nl=0(2ll)mlNlk=0(lk)(1)k4lk(k+l)sl+k,k=Nl=0l(2ll)mlNlk=0((lk)+(l1k1))(1)k4lksl+k,k=N/2l=0l(2ll)mllk=0((lk)+(l1k1))(1)k4lksl+k,k+N/2<lNl(2ll)mlNlk=0((lk)+(l1k1))(1)k4lksl+k,k,

    where we consider (x1) as 0.

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

    |Nlk=0(lk)(1)k4lksl+k,k|lk=0(lk)4lksl+k,klk=0(lk)4lk(2k+1)4kl(k+ll)l(2l+1)4llk=0(lk)(l+kk)=l(2l+1)4lPl(3),

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

    |N/2<lN(2ll)mlNlk=0(lk)(1)k4lksl+k,k|N/2<lNl(2l+1)(16m)lPl(3)l>N/2l(2l+1)Pl(3)(16m)l

    and

    |N/2<lNl(2ll)mlNlk=0((lk)+(l1k1))(1)k4lksl+k,k|N/2<lNl4lmllk=02(lk)4lksl+k,k
    N/2<lN2l2(2l+1)(16m)lPl(3)2l>N/2l2(2l+1)Pl(3)(16m)l.

    Recall that

    Pl(3)=Tl(3,2)(3+22)l+1/2242lπ  as l+.

    As |m|94, we have |m|>16(3+22)93.255 and hence

    l=0l2(2l+1)Pl(3)(16m)l

    converges. Thus

    limN+l>N/2l(2l+1)Pl(3)(16m)l=0=limN+l>N/2l2(2l+1)Pl(3)(16m)l

    and hence by the above we have

    n=0Sn(4,m)mn=l=0(2ll)mllk=0(lk)(1)k4lksl+k,k

    and

    n=0nSn(4,m)mn=l=0l(2ll)mllk=0((lk)+(l1k1))(1)k4lksl+k,k.

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

    n=0Sn(4,m)mn=n=0(2nn)mnfn (2.24)

    and

    n=0nSn(4,m)mn=n=0n(2nn)mn(fn+tn). (2.25)

    In view of (2.25) and (2.20),

    (m+16)n=0nSn(4,m)mn=n=1n(2nn)mn1(fn+tn)+16n=0n(2nn)mn(fn+tn)=n=0(n+1)(2n+2n+1)(fn+1+tn+1)+16n(2nn)(fn+tn)mn=2n=0(2nn)mn((2n+1)(fn+1+tn+1)+8n(fn+tn))
    =2n=0(2nn)mn(2(2n+1)fn+1+4(n1)fn)=2n=0(n+1)(2n+2n+1)fn+1mn+8n=0(n1)(2nn)fnmn=2n=0n(2nn)fnmn1+8n=0(n1)(2nn)fnmn=2n=0((m+4)n4)(2nn)fnmn.

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

    Proof of Theorem 1.3. Let a,b,mZ with |m|6. Since

    4nTn(1,m)=n/2k=0(n2k)(2kk)4n2k(16m)k=Tn(4,16m)

    for any nN, we have 4nSn(1,m)=Sn(4,16m) for all nN. Thus, in light of Theorem 2.6,

    n=0(an+b)Sn(1,m)(4m)n=n=0(an+b)Sn(4,16m)(16m)n=11616mn=0(2a(416m)n8a+(1616m)b)(2nn)fn(16m)n=12(m1)n=0(a(4m1)n+a+2b(m1))(2nn)fn(16m)n.

    Therefore

    k=07k+324kSk(1,6)=52k=05k+196k(2kk)fk,k=012k+5(28)kSk(1,7)=3k=09k+2(112)k(2kk)fk,k=084k+2980kSk(1,20)=27k=06k+1320k(2kk)fk,k=03k+1(100)kSk(1,25)=116k=099k+17(400)k(2kk)fk,k=0228k+67224kSk(1,56)=5k=090k+13896k(2kk)fk,k=0399k+101(676)kSk(1,169)=1516k=0855k+109(2704)k(2kk)fk,k=02604k+5632600kSk(1,650)=51k=0102k+1110400k(2kk)fk,k=039468k+7817(6076)kSk(1,1519)=135k=0585k+58(24304)k(2kk)fk,k=041667k+78799800kSk(1,2450)=2972k=0561k+5339200k(2kk)fk,k=074613k+10711(5302)kSk(1,2652)=2332k=0207621k+14903(10602)k(2kk)fk.

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

    k=05k+196k(2kk)fk=32π,  k=09k+2(112)k(2kk)fk=27π,k=06k+1320k(2kk)fk=8159π,  k=099k+17(400)k(2kk)fk=50π,k=090k+13896k(2kk)fk=167π,  k=0855k+109(2704)k(2kk)fk=338π,k=0102k+1110400k(2kk)fk=50399π,  k=0585k+58(24304)k(2kk)fk=98313π,k=0561k+5339200k(2kk)fk=1225618π,  k=0207621k+14903(10602)k(2kk)fk=1404503π.

    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:

    k=050k+1(256)k(2kk)(2kk+1)Tk(1,16)=83π,(1)
    k=0(100k24k7)(2kk)2Tk(1,16)(2k1)2(256)k=24π,(1)
    k=030k+23(1024)k(2kk)(2kk+1)Tk(34,1)=203π,(2)
    k=0(36k212k+1)(2kk)2Tk(34,1)(2k1)2(1024)k=6π,(2)
    k=0110k+1034096k(2kk)(2kk+1)Tk(194,1)=304π,(3)
    k=0(20k2+28k11)(2kk)2Tk(194,1)(2k1)24096k=6π,(3)
    k=0238k+2634096k(2kk)(2kk+1)Tk(62,1)=11233π,(4)
    k=0(44k2+4k5)(2kk)2Tk(62,1)(2k1)24096k=43π,(4)
    k=06k+1256k(2kk)2Tk(8,2)=2π8+62,(5)
    k=02k+3256k(2kk)(2kk+1)Tk(8,2)=68+6216423π,(5)
    k=0(4k2+2k1)(2kk)2Tk(8,2)(2k1)2256k=3424π.(5)

    Remark 3.1. For each kN, we have

    ((1+λ0λ1)k+λ0)Ck=(k+λ0)(2kk)(k+λ1)(2kk+1)

    since (2kk)=(k+1)Ck and (2kk+1)=kCk. Thus, for example, [40,(I1)] and (I1) together imply that

    k=026k+5(256)k(2kk)CkTk(1,16)=16π,

    and (I5) and (I5) imply that

    k=02k1256k(2kk)CkTk(8,2)=4π(8+62442).

    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

    p1k=030k+23(1024)k(2kk)(2kk+1)Tk(34,1)p3(21(2p)10(1p)11) (mod p2)

    and

    p1k=02k+1(1024)k(2kk)CkTk(34,1)p3(23(2p)+4(1p)) (mod p2).

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

    12n/2+1n(2nn)n1k=0(6k+1)(2kk)2Tk(8,2)256n1kZ+

    and

    1(2n2n1)n1k=0(12k4k2)(2kk)2Tk(8,2)(2k1)2256kZ+

    for each n=2,3,, and that for any prime p1 (mod 4) with p=x2+4y2 (x,yZ) we have

    p1k=0(2kk)2Tk(8,2)256k{(1)y/2(4x22p) (mod p2)if p1 (mod 8),(1)(xy1)/28xy (mod p2)if p5 (mod 8),

    and

    p1k=0(4k2+2k1)(2kk)2Tk(8,2)(2k1)2256k0 (mod p2).

    By [40,Theorem 5.1], we have

    p1k=0(2kk)2Tk(8,2)256k0 (mod p2)

    for any prime p3 (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:

    k=03k+4972k(2kk+1)(3kk)Tk(18,6)=63340π,(1)
    k=091k+107103k(2kk+1)(3kk)Tk(10,1)=275318π,(2)
    k=0195k+83183k(2kk+1)(3kk)Tk(198,1)=9423310π,(3)
    k=0483k419303k(2kk+1)(3kk)Tk(970,1)=65503π,(4)
    k=0666k+757303k(2kk+1)(3kk)Tk(730,729)=347534π,(5)
    k=08427573k+84421071023k(2kk+1)(3kk)Tk(102,1)=125137620π,(6)
    k=0959982231k+9604225031983k(2kk+1)(3kk)Tk(198,1)=5335011320π,(7)
    k=099k+1243k(2kk+1)(3kk)Tk(26,729)=16(289156453)15π,(10)
    k=045k+1(5400)k(2kk+1)(3kk)Tk(70,3645)=3453157156π,(11)
    k=0252k1(13500)k(2kk+1)(3kk)Tk(40,1458)=25(121238596)24π,(12)
    k=09k+2(675)k(2kk)(3kk)Tk(15,5)=7158π,(13)
    k=045k+31(675)k(2kk+1)(3kk)Tk(15,5)=19158π,(13)
    k=039k+7(1944)k(2kk)(3kk)Tk(18,3)=93π,(14)
    k=0312k+263(1944)k(2kk+1)(3kk)Tk(18,3)=4532π.(14)

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

    p1k=039k+7(1944)k(2kk)(3kk)Tk(18,3)p2(13(p3)+1) (mod p2)

    and that

    p1k=0(2kk)(3kk)Tk(18,3)(1944)k{4x22p (mod p2)if (1p)=(p3)=(p7)=1  p=x2+21y2,2p2x2 (mod p2)if (1p)=(p3)=1, (p7)=1  2p=x2+21y2,12x22p (mod p2)if (1p)=(p7)=1, (p3)=1  p=3x2+7y2,2p6x2 (mod p2)if (1p)=1, (p3)=(p7)=1  2p=3x2+7y2,0 (mod p2)if (21p)=1,

    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:

    k=017k+18662k(2kk+1)(4k2k)Tk(52,1)=773312π,(1)
    k=04k+3(962)k(2kk+1)(4k2k)Tk(110,1)=63π,(2)
    k=08k+91122k(2kk+1)(4k2k)Tk(98,1)=15421135π,(3)
    k=03568k+40272642k(2kk+1)(4k2k)Tk(257,256)=8696610π,(4)
    k=0144k+1(1682)k(2kk+1)(4k2k)Tk(7,4096)=7(174542778210)120π,(5)
    k=03496k+37093362k(2kk+1)(4k2k)Tk(322,1)=1827π,(6)
    k=0286k+2293362k(2kk+1)(4k2k)Tk(1442,1)=111321020π,(7)
    k=08426k+86339122k(2kk+1)(4k2k)Tk(898,1)=70311420π,(8)
    k=01608k+799122k(2kk+1)(4k2k)Tk(12098,1)=67849399105π,(9)
    k=0134328722k+134635283104162k(2kk+1)(4k2k)Tk(10402,1)=939614344π,(10)

    and

    k=039600310408k+39624469807392162k(2kk+1)(4k2k)Tk(39202,1)=1334161817π.(12)

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

    Conjecture 3.4. We have the following identities:

    k=0(356k2+288k+7)(2kk)2T2k(7,1)(k+1)(2k1)(482)k=3043π,(1)
    k=0(172k2+141k1)(2kk)2T2k(62,1)(k+1)(2k1)(4802)k=803π,(2)
    k=0(782k2+771k+19)(2kk)2T2k(322,1)(k+1)(2k1)(57602)k=90π,(3)
    k=0(34k2+45k+5)(2kk)2T2k(10,1)(k+1)(2k1)962k=2023π,(4)
    k=0(106k2+193k+27)(2kk)2T2k(38,1)(k+1)(2k1)2402k=106π,(5)
    k=0(214166k2+221463k+7227)(2kk)2T2k(198,1)(k+1)(2k1)392002k=92406π,(6)
    k=0(112k2+126k+9)(2kk)2T2k(18,1)(k+1)(2k1)3202k=615π,(7)
    k=0(926k2+995k+55)(2kk)2T2k(30,1)(k+1)(2k1)8962k=607π,(8)
    k=0(1136k2+2962k+503)(2kk)2T2k(110,1)(k+1)(2k1)244k=907π,(9)
    k=0(5488k2+8414k+901)(2kk)2T2k(322,1)(k+1)(2k1)484k=2947π,(10)
    k=0(170k2+193k+11)(2kk)2T2k(198,1)(k+1)(2k1)28002k=614π,(11)
    k=0(104386k2+108613k+4097)(2kk)2T2k(102,1)(k+1)(2k1)104002k=204039π,(12)
    k=0(7880k2+8217k+259)(2kk)2T2k(1298,1)(k+1)(2k1)468002k=14426π,(13)
    k=0(6152k2+45391k+9989)(2kk)2T2k(1298,1)(k+1)(2k1)56162k=6633π,(14)
    k=0(147178k2+2018049k+471431)(2kk)2T2k(4898,1)(k+1)(2k1)204002k=374051π,(15)
    k=0(1979224k2+5771627k+991993)(2kk)2T2k(5778,1)(k+1)(2k1)288802k=7387210π,(16)
    k=0(233656k2+239993k+5827)(2kk)2T2k(5778,1)(k+1)(2k1)4392802k=408019π,(17)
    k=0(5890798k2+32372979k+6727511)(2kk)2T2k(54758,1)(k+1)(2k1)2433602k=600704959π,(18)
    k=0(148k2+272k+43)(2kk)2T2k(10,2)(k+1)(2k1)4608k=286π,(19)
    k=0(3332k2+17056k+3599)(2kk)2T2k(238,14)(k+1)(2k1)1161216k=7442π,(20)
    k=0(11511872k2+10794676k+72929)(2kk)2T2k(9918,19)(k+1)(2k1)(16629048064)k=3903547π.(21)

    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 nZ+. If (3p)=1, then

    pn1k=066k+17(21133)kTk(10,112)3p(2p)n1k=066k+17(21133)kTk(10,112)3

    divided by (pn)2 is a p-adic integer. If p5, then

    pn1k=0126k+31(80)3kTk(22,212)3p(5p)n1k=0126k+31(80)3kTk(22,212)3

    divided by (pn)2 is a p-adic integer. If (7p)=1 but p3, then

    pn1k=03990k+1147(288)3kTk(62,952)3p(2p)n1k=03990k+1147(288)3kTk(62,952)3

    divided by (pn)2 is a p-adic integer. If p±1 (mod 8) but p7, then

    pn1k=024k+5282k(2kk)Tk(4,9)2p(p3)n1k=024k+5282k(2kk)Tk(4,9)2

    divided by (pn)2 is a p-adic integer. If (6p)=1 but p7,31, then

    pn1k=02800512k+4352574342k(2kk)Tk(73,576)2pn1k=02800512k+4352574342k(2kk)Tk(73,576)2

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

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

    Conjecture 3.6. We have

    k=040k+13(50)kTk(4,1)Tk(1,1)2=55159π,(1)
    k=01435k+1133240kTk(7,1)Tk(10,10)2=14525π,(2)
    k=0840k+197(2430)kTk(8,1)Tk(5,5)2=189152π,(3)
    k=039480k+7321(29700)kTk(14,1)Tk(11,11)2=67955π.(4)

    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 nZ+, we have

    1nn1k=0(40k+13)(1)k50n1kTk(4,1)Tk(1,1)2Z+, (3.1)

    and this number is odd if and only if n is a power of two (i.e., n{2a: aN}).

    (ⅱ) Let p2,5 be a prime. Then

    p1k=040k+13(50)kTk(4,1)Tk(1,1)2p3(12+5(3p)+22(15p)) (mod p2). (3.2)

    If (3p)=(5p)=1, then

    1(pn)2(pn1k=040k+13(50)kTk(4,1)Tk(1,1)2pn1k=040k+13(50)kTk(4,1)Tk(1,1)2)Zp (3.3)

    for all nZ+.

    (ⅲ) Let p2,5 be a prime. Then

    p1k=0Tk(4,1)Tk(1,1)2(50)k{4x22p (mod p2)if p1,9 (mod 20)  p=x2+5y2 (x,yZ),2x22p (mod p2)if p3,7 (mod 20)  2p=x2+5y2 (x,yZ),0 (mod p2)if (5p)=1. (3.4)

    Remark 3.4. The imaginary quadratic field Q(5) has class number two.

    Conjecture 3.8. (ⅰ) For any nZ+, we have

    1nn1k=0(40k+27)(6)n1kTk(4,1)Tk(1,1)2Z, (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

    p1k=040k+27(6)kTk(4,1)Tk(1,1)2p9(55(5p)+198(3p)10) (mod p2). (3.6)

    If (3p)=(5p)=1, then

    1(pn)2(pn1k=040k+27(6)kTk(4,1)Tk(1,1)2pn1k=040k+27(6)kTk(4,1)Tk(1,1)2)Zp (3.7)

    for all nZ+.

    (ⅲ) Let p>5 be a prime. Then

    (p3)p1k=0Tk(4,1)Tk(1,1)2(6)k{4x22p (mod p2)if p1,9 (mod 20)  p=x2+5y2 (x,yZ),2p2x2 (mod p2)if p3,7 (mod 20)  2p=x2+5y2 (x,yZ),0 (mod p2)if (5p)=1. (3.8)

    Remark 3.5. This conjecture can be viewed as the dual of Conjecture 3.7. Note that the series k=0(40k+27(6)kTk(4,1)Tk(1,1)2 diverges.

    Conjecture 3.9. (ⅰ) For each nZ+, we have

    1n10n1n1k=0(1435k+113)3240n1kTk(7,1)Tk(10,10)2Z+. (3.9)

    (ⅱ) Let p>3 be a prime. Then

    p1k=01435k+1133240kTk(7,1)Tk(10,10)2p9(2420(5p)+105(5p)1508) (mod p2). (3.10)

    If p1,9 (mod 20), then

    pn1k=01435k+1133240kTk(7,1)Tk(10,10)2pn1k=01435k+1133240kTk(7,1)Tk(10,10)2 (3.11)

    divided by (pn)2 is a p-adic integer for each nZ+.

    (ⅲ) Let p>5 be a prime. Then

    p1k=0Tk(7,1)Tk(10,10)23240k{4x22p (mod p2)if p1,4 (mod 15)  p=x2+15y2 (x,yZ),12x22p (mod p2)if p2,8 (mod 15)  p=3x2+5y2 (x,yZ),0 (mod p2)if (15p)=1. (3.12)

    Remark 3.6. The imaginary quadratic field Q(15) has class number two.

    Conjecture 3.10. (ⅰ) For each nZ+, we have

    32n10n1n1k=0(1435k+1322)50n1kTk(7,1)Tk(10,10)2Z+. (3.13)

    (ⅱ) Let p>5 be a prime. Then

    p1k=01435k+132250kTk(7,1)Tk(10,10)2p3(3432(5p)+968(1p)434) (mod p2). (3.14)

    If p1,9 (mod 20), then

    pn1k=01435k+132250kTk(7,1)Tk(10,10)2pn1k=01435k+132250kTk(7,1)Tk(10,10)2 (3.15)

    divided by (pn)2 is a p-adic integer for each nZ+.

    (ⅲ) Let p>5 be a prime. Then

    p1k=0Tk(7,1)Tk(10,10)250k{4x22p (mod p2)if p1,4 (mod 15)  p=x2+15y2 (x,yZ),2p12x2 (mod p2)if p2,8 (mod 15)  p=3x2+5y2 (x,yZ),0 (mod p2)if (15p)=1. (3.16)

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

    k=01435k+132250kTk(7,1)Tk(10,10)2

    diverges.

    Conjecture 3.11. (ⅰ) For each nZ+, we have

    1n5n1n1k=0(840k+197)(1)k2430n1kTk(8,1)Tk(5,5)2Z+. (3.17)

    (ⅱ) Let p>3 be a prime. Then

    p1k=0840k+197(2430)kTk(8,1)Tk(5,5)2p(140(15p)+5(15p)+52) (mod p2). (3.18)

    If (1p)=(15p)=1, then

    pn1k=0840k+197(2430)kTk(8,1)Tk(5,5)2pn1k=0840k+197(2430)kTk(8,1)Tk(5,5)2 (3.19)

    divided by (pn)2 is an p-adic integer for any nZ+.

    (ⅲ Let p>7 be a prime. Then

    p1k=0Tk(8,1)Tk(5,5)2(2430)k{4x22p (mod p2)if (1p)=(p3)=(p5)=(p7)=1, p=x2+105y2,2x22p (mod p2)if (1p)=(p7)=1, (p3)=(p5)=1, 2p=x2+105y2,12x22p (mod p2)if (1p)=(p3)=(p5)=(p7)=1, p=3x2+35y2,6x22p (mod p2)if (1p)=(p7)=1, (p3)=(p5)=1, 2p=3x2+35y2,2p20x2 (mod p2)if (1p)=(p5)=1, (p5)=(p7)=1, p=5x2+21y2,2p10x2 (mod p2)if (1p)=(p3)=1, (p5)=(p7)=1, 2p=5x2+21y2,28x22p (mod p2)if (1p)=(p5)=1, (p3)=(p7)=1, p=7x2+15y2,14x22p (mod p2)if (1p)=(p3)=1, (p5)=(p7)=1, 2p=7x2+15y2,0 (mod p2)if (105p)=1, (3.20)

    where x and y are integers.

    Remark 3.8. Note that the imaginary quadratic field Q(105) has class number 8.

    Conjecture 3.12. (ⅰ) For each nZ+, we have

    1nn1k=0(39480k+7321)(1)k29700n1kTk(14,1)Tk(11,11)2Z+, (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

    p1k=039480k+7321(29700)kTk(14,1)Tk(11,11)2p(5738(5p)+70(3p)+1513) (mod p2). (3.22)

    If (3p)=(5p)=1, then

    pn1k=039480k+7321(29700)kTk(14,1)Tk(11,11)2pn1k=039480k+7321(29700)kTk(14,1)Tk(11,11)2 (3.23)

    divided by (pn)2 is a p-adic integer for each nZ+.

    (ⅲ) Let p>5 be a prime with p11. Then

    p1k=0Tk(14,1)Tk(11,11)2(29700)k{4x22p (mod p2)if (1p)=(p3)=(p5)=(p11)=1, p=x2+165y2,2x22p (mod p2)if (1p)=(p3)=(p5)=(p11)=1, 2p=x2+165y2,12x22p (mod p2)if (1p)=(p5)=1, (p3)=(p11)=1, p=3x2+55y2,6x22p (mod p2)if (1p)=(p5)=1, (p3)=(p11)=1, 2p=3x2+55y2,2p20x2 (mod p2)if (1p)=(p11)=1, (p3)=(p5)=1, p=5x2+33y2,2p10x2 (mod p2)if (1p)=(p11)=1, (p3)=(p5)=1, 2p=5x2+33y2,44x22p (mod p2)if (1p)=(p3)=1, (p5)=(p11)=1, p=11x2+15y2,22x22p (mod p2)if (1p)=(p3)=1, (p5)=(p11)=1, 2p=11x2+15y2,0 (mod p2)if (165p)=1, (3.24)

    where x and y are integers.

    Remark 3.9. Note that the imaginary quadratic field Q(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 nZ+, we have

    1nn1k=0(7k+3)Sk(1,6)24n1kZ+. (4.1)

    (ⅱ) Let p>3 be a prime. Then

    p1k=07k+324kSk(1,6)p2(5(2p)+(6p)) (mod p2). (4.2)

    If p1 (mod 3), then

    1(pn)2(pn1k=07k+324kSk(1,6)p(2p)n1k=07k+324kSk(1,6))Zp (4.3)

    for all nZ+.

    (ⅲ) For any prime p>3, we have

    p1k=0Sk(1,6)24k{(p3)(4x22p) (mod p2)if p1,3 (mod 8)  p=x2+2y2 (x,yZ),0 (mod p2)if p5,7 (mod 8). (4.4)

    Conjecture 4.2. (ⅰ) For any nZ+, we have

    1nn1k=0(12k+5)Sk(1,7)(1)k28n1kZ+, (4.5)

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

    (ⅱ) Let p7 be an odd prime. Then

    p1k=012k+5(28)kSk(1,7)5p(p7) (mod p2), (4.6)

    and moreover

    1(pn)2(pn1k=012k+5(28)kSk(1,7)p(p7)n1k=012k+5(28)kSk(1,7))Zp (4.7)

    for all nZ+.

    (ⅲ) For any prime p2,7, we have

    p1k=0Sk(1,7)(28)k{4x22p (mod p2)if (1p)=(p3)=(p7)=1  p=x2+21y2,2x22p (mod p2)if (1p)=(p3)=1, (p7)=1  2p=x2+21y2,2p12x2 (mod p2)if (1p)=(p7)=1, (p3)=1  p=3x2+7y2,2p6x2 (mod p2)if (1p)=1, (p3)=(p7)=1  2p=3x2+7y2,0 (mod p2)if (21p)=1, (4.8)

    where x and y are integers.

    Conjecture 4.3. (ⅰ) For any nZ+, we have

    1nn1k=0(84k+29)Sk(1,20)80n1kZ+, (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 p5. Then

    p1k=084k+2980kSk(1,20)p(2(5p)+27(15p)) (mod p2). (4.10)

    If p1 (mod 3), then

    1(pn)2(pn1k=084k+2980kSk(1,20)p(p5)n1k=084k+2980kSk(1,20))Zp (4.11)

    for all nZ+.

    (ⅲ) For any prime p2,5, we have

    p1k=0Sk(1,20)80k{4x22p (mod p2)if (2p)=(p3)=(p5)=1  p=x2+30y2,8x22p (mod p2)if (2p)=1, (p3)=(p5)=1  p=2x2+15y2,2p12x2 (mod p2)if (p3)=1, (2p)=(p5)=1  p=3x2+10y2,20x22p (mod p2)if (p5)=1, (2p)=(p3)=1  p=5x2+6y2,0 (mod p2)if (30p)=1, (4.12)

    where x and y are integers.

    Conjecture 4.4. (ⅰ) For any nZ+, we have

    1nn1k=0(3k+1)(1)k100n1kSk(1,25)Z+. (4.13)

    (ⅱ) Let p5 be an odd prime. Then

    1(pn)2(pn1k=03k+1(100)kSk(1,25)p(1p)n1k=03k+1(100)kSk(1,25))Zp (4.14)

    for all nZ+.

    (ⅲ) For any prime p>3 with p11, we have

    p1k=0Sk(1,25)(100)k{4x22p (mod p2)if (1p)=(p3)=(p11)=1  p=x2+33y2,2x22p (mod p2)if (1p)=1, (p3)=(p11)=1  2p=x2+33y2,2p12x2 (mod p2)if (p11)=1, (1p)=(p3)=1  p=3x2+11y2,2p6x2 (mod p2)if (p3)=1, (1p)=(p11)=1  2p=3x2+11y2,0 (mod p2)if (33p)=1, (4.15)

    where x and y are integers.

    Conjecture 4.5. (ⅰ) For any nZ+, we have

    1nn1k=0(228k+67)Sk(1,56)224n1kZ+, (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 p7. Then

    p1k=0228k+67224kSk(1,56)p(65(7p)+2(14p)) (mod p2). (4.17)

    If p1,3 (mod 8), then

    1(pn)2(pn1k=0228k+67224kSk(1,56)p(p7)n1k=0228k+67224kSk(1,56))Zp (4.18)

    for all nZ+.

    (ⅲ) For any prime p2,7, we have

    p1k=0Sk(1,56)224k{4x22p (mod p2)if (2p)=(p3)=(p7)=1  p=x2+42y2,8x22p (mod p2)if (p7)=1, (2p)=(p3)=1  p=2x2+21y2,2p12x2 (mod p2)if (2p)=1, (p3)=(p7)=1  p=3x2+14y2,2p24x2 (mod p2)if (p5)=1, (2p)=(p3)=1  p=6x2+7y2,0 (mod p2)if (42p)=1, (4.19)

    where x and y are integers.

    Conjecture 4.6. (ⅰ) For any nZ+, we have

    1nn1k=0(399k+101)(1)k676n1kSk(1,169)Z+. (4.20)

    (ⅱ) Let p13 be an odd prime. Then

    pn1k=0399k+101(676)kSk(1,169)p(1p)n1k=0399k+101(676)kSk(1,169) (4.21)

    divided by (pn)2 is a p-adic integer for any nZ+.

    (ⅲ) For any prime p>3 with p19, we have

    p1k=0Sk(1,169)(676)k{4x22p (mod p2)if (1p)=(p3)=(p19)=1  p=x2+57y2,2x22p (mod p2)if (1p)=1, (p3)=(p19)=1  2p=x2+57y2,2p12x2 (mod p2)if (p3)=1, (1p)=(p19)=1  p=3x2+19y2,2p6x2 (mod p2)if (p19)=1, (1p)=(p3)=1  2p=3x2+19y2,0 (mod p2)if (57p)=1, (4.22)

    where x and y are integers.

    Conjecture 4.7. (ⅰ) For any nZ+, we have

    1nn1k=0(2604k+563)Sk(1,650)2600n1kZ+, (4.23)

    and this number is odd if and only if n{2a: aN}.

    (ⅱ) Let p be an odd prime with p5,13. Then

    p1k=02604k+5632600kSk(1,650)p(561(39p)+2(26p)) (mod p2). (4.24)

    If (6p)=1, then

    pn1k=02604k+5632600kSk(1,650)p(26p)n1k=02604k+5632600kSk(1,650) (4.25)

    divided by (pn)2 is a p-adic integer for any nZ+.

    (ⅲ)For any odd prime p5,13, we have

    p1k=0Sk(1,650)2600k{4x22p (mod p2)if (2p)=(p3)=(p13)=1  p=x2+78y2,8x22p (mod p2)if (2p)=1, (p3)=(p13)=1  p=2x2+39y2,2p12x2 (mod p2)if (p13)=1, (2p)=(p3)=1  p=3x2+26y2,2p24x2 (mod p2)if (p3)=1, (2p)=(p13)=1  p=6x2+13y2,0 (mod p2)if (78p)=1, (4.26)

    where x and y are integers.

    Conjecture 4.8. (ⅰ) For any nZ+, we have

    1nn1k=0(39468k+7817)(1)k6076n1kSk(1,1519)Z+, (4.27)

    and this number is odd if and only if n{2a: aN}.

    (ⅱ) Let p7,31 be an odd prime. Then

    pn1k=039468k+7817(6076)kSk(1,1519)p(31p)n1k=039468k+7817(6076)kSk(1,1519) (4.28)

    divided by (pn)2 is a p-adic integer for any nZ+.

    (ⅲ) For any prime p>3 with p7,31, we have

    p1k=0Sk(1,1519)(6076)k{4x22p (mod p2)if (1p)=(p3)=(p31)=1  p=x2+93y2,2x22p (mod p2)if (p31)=1, (1p)=(p3)=1  2p=x2+93y2,2p12x2 (mod p2)if (p3)=1, (1p)=(p31)=1  p=3x2+31y2,2p6x2 (mod p2)if (1p)=1, (p3)=(p31)=1  2p=3x2+31y2,0 (mod p2)if (93p)=1, (4.29)

    where x and y are integers.

    Conjecture 4.9. (ⅰ) For any nZ+, we have

    1nn1k=0(41667k+7879)9800n1kSk(1,2450)Z+. (4.30)

    (ⅱ) Let p5,7 be an odd prime. Then

    p1k=041667k+78799800kSk(1,2450)p2(15741(6p)+17(2p)) (mod p2). (4.31)

    If p1 (mod 3), then

    pn1k=041667k+78799800kSk(1,2450)p(2p)n1k=041667k+78799800kSk(1,2450) (4.32)

    divided by (pn)2 is a p-adic integer for any nZ+.

    (ⅲ) For any prime p>7 with p17, we have

    p1k=0Sk(1,2450)9800k{4x22p (mod p2)if (2p)=(p3)=(p17)=1  p=x2+102y2,8x22p (mod p2)if (p17)=1, (2p)=(p3)=1  p=2x2+51y2,2p12x2 (mod p2)if (p3)=1, (2p)=(p17)=1  p=3x2+34y2,2p24x2 (mod p2)if (2p)=1, (p3)=(p17)=1  p=6x2+17y2,0 (mod p2)if (102p)=1, (4.33)

    where x and y are integers.

    Conjecture 4.10. (ⅰ) For any nZ+, we have

    1nn1k=0(74613k+10711)(1)k5302(n1k)Sk(1,2652)Z+. (4.34)

    () Let p5,53 be an odd prime. Then

    pn1k=074613k+10711(5302)kSk(1,2652)p(1p)n1k=074613k+10711(5302)kSk(1,2652) (4.35)

    divided by (pn)2 is a p-adic integer for any nZ+.

    () For any prime p>5 with p59, we have

    p1k=0Sk(1,2652)(5302)k{4x22p (mod p2)if (1p)=(p3)=(p59)=1  p=x2+177y2,2x22p (mod p2)if (1p)=1, (p3)=(p59)=1  2p=x2+177y2,2p12x2 (mod p2)if (p59)=1, (1p)=(p3)=1  p=3x2+59y2,2p6x2 (mod p2)if (p3)=1, (1p)=(p59)=1  2p=3x2+59y2,0 (mod p2)if (177p)=1, (4.36)

    where x and y are integers.

    Conjecture 4.11. For any odd prime p,

    p1k=0Sk(4)k{4x22p (mod p2)if 12p1  p=x2+y2 (x,yZ  3x),4xy (mod p2)if 12p5  p=x2+y2 (x,yZ  3xy),0 (mod p2)if p3 (mod 4). (4.37)

    Also, for any prime p1 (mod 4) we have

    p1k=0(8k+5)Sk(4)k4p (mod p2). (4.38)

    Conjecture 4.12. () For any nZ+, we have

    1nn1k=0(4k+3)4n1kSk(1,1)Z, (4.39)

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

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

    1(pn)2(pn1k=04k+34kSk(1,1)pn1k=04k+34kSk(1,1))Zp. (4.40)

    () Let p be an odd prime. Then

    p1k=0Sk(1,1)4k{4x22p (mod p2)if p1,9 (mod 20)  p=x2+5y2 (x,yZ),2x22p (mod p2)if p3,7 (mod 20)  2p=x2+5y2 (x,yZ),0 (mod p2)if (5p)=1. (4.41)

    Conjecture 4.13. () For any nZ+, we have

    1nn1k=0(33k+25)Sk(1,6)(6)n1kZ, (4.42)

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

    () Let p>3 be a prime. Then

    p1k=033k+25(6)kSk(1,6)p(3510(3p)) (mod p2). (4.43)

    If p±1 (mod 12), then

    1(pn)2(pn1k=033k+25(6)kSk(1,6)pn1k=033k+25(6)kSk(1,6))Zp (4.44)

    for all nZ+.

    () For any prime p>3, we have

    p1k=0Sk(1,6)(6)k{(1p)(4x22p) (mod p2)if p=x2+3y2 (x,yZ),0 (mod p2)if p2 (mod 3). (4.45)

    Conjecture 4.14. () For any nZ+, we have

    n  | n1k=0(18k+13)Sk(2,9)8n1k. (4.46)

    () Let p be an odd prime. Then

    p1k=018k+138kSk(2,9)p(1+12(p3)) (mod p2). (4.47)

    If p1 (mod 3), then

    1(pn)2(pn1k=018k+138kSk(2,9)pn1k=018k+138kSk(2,9))Zp (4.48)

    for all nZ+.

    () For any prime p>3, we have

    p1k=0Sk(1,2)8k(p3)p1k=0Sk(2,9)8k{4x22p (mod p2)if p1,7 (mod 24)  p=x2+6y2 8x22p (mod p2)if p5,11 (mod 24)  p=2x2+3y2,0 (mod p2)if (6p)=1, (4.49)

    where x and y are integers.

    Conjecture 4.15. Let p be an odd prime with p5. Then

    p1k=0Sk(3,1)4k{4x22p (mod p2)if p1,9 (mod 20)  p=x2+5y2, 2x22p (mod p2)if p3,7 (mod 20)  2p=x2+5y2,0 (mod p2)if p11,13,17,19 (mod 20), (4.50)

    where x and y are integers. If (5p)=1, then

    p1k=040k+294kSk(3,1)18p (mod p2).

    Remark 4.1. We also have some similar conjectures involving

    p1k=0Sk(5,4)4k, p1k=0Sk(4,5)4k, p1k=0Sk(7,6)6k,p1k=0Sk(10,2)32k, p1k=0Sk(14,9)72k, p1k=0Sk(19,9)36k

    modulo p2, 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 m0 (mod p), we have

    p1k=0Sk(4,m)mkp1k=0(2kk)fkmk (mod p2). (4.51)

    and

    m+162p1k=0kSk(4,m)mkp1k=0((m+4)k4)(2kk)fkmk4p(mp) (mod p2). (4.52)

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

    The numbers

    Zn:=nk=0(nk)(2kk)(2(nk)nk)  (n=0,1,2,)

    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 nN the number 2nZn coincides with the so-called CLF (Catalan-Larcombe-French) number

    Pn:=2nn/2k=1(n2k)(2kk)24n2k=nk=0(2kk)2(2(nk)nk)2(nk).

    Let p be an odd prime. For any k=0,,p1, we have

    Pk(1p)128kPp1k (mod p)

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

    Zk=Pk2k(1p)64k(2p1kZp1k)(1p)32kZp1k (mod p).

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

    p1k=0ZkTk(b,c)mk(4cb2p)p1k=0(32(b24c)m)kZp1kTp1k(b,c)(4cb2p)p1k=0ZkTk(b,c)(32(b24c)/m)k (mod p)

    for any b,c,mZ with p(b24c)m.

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

    k=0(15k+426)ZkPk(246152)(46103)k=6π(7+36).

    Via our congruence approach (including Conjecture 1.4), we find 24 rational series for 1/π involving Tn(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/π.

    k=15k+132kTkZk=8(2+5)3π, (5.1)
    k=021k+5(252)kTk(1,16)Zk=67π, (5.2)
    k=03k+136kTk(1,2)Zk=3π, (5.3)
    k=0k192kTk(14,1)Zk=83π, (5.4)
    k=030k+11(192)kTk(14,1)Zk=12π, (5.5)
    k=015k+1480kTk(22,1)Zk=610π, (5.6)
    k=07k+2(672)kTk(26,1)Zk=2213π, (5.7)
    k=021k+21152kTk(34,1)Zk=18π, (5.8)
    k=030k7640kTk(62,1)Zk=160π, (5.9)
    k=0195k+34(9600)kTk(98,1)Zk=80π, (5.10)
    k=0195k+2211232kTk(106,1)Zk=2713π, (5.11)
    k=042k+17(1440)kTk(142,1)Zk=335π, (5.12)
    k=02k11792kTk(194,1)Zk=563π, (5.13)
    k=01785k+254(37632)kTk(194,1)Zk=672π, (5.14)
    k=0210k+2340800kTk(202,1)Zk=1534π, (5.15)
    k=0210k14608kTk(254,1)Zk=288π, (5.16)
    k=021k55600kTk(502,1)Zk=1052π, (5.17)
    k=07410k+1849(36992)kTk(1154,1)Zk=2992π, (5.18)
    k=01326k+10157760kTk(1442,1)Zk=20145π, (5.19)
    k=078k13120800kTk(2498,1)Zk=2600π, (5.20)
    k=062985k+11363(394272)kTk(5474,1)Zk=765910π, (5.21)
    k=0358530k+33883486720kTk(6082,1)Zk=176280π, (5.22)
    k=0510k152378400kTk(9602,1)Zk=33320π, (5.23)
    k=0570k45793600kTk(10402,1)Zk=159013π. (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 nZ+, we have

    n | n1k=0(5k+1)TkZk32n1k. (5.25)

    () Let p be an odd prime with p5. Then

    p1k=05k+132kTkZkp3(5(5p)2(1p)) (mod p2). (5.26)

    If p±1 (mod 5), then

    1(pn)2(pn1k=05k+132kTkZkp(1p)n1k=05k+132kTkZk)Zp (5.27)

    for all nZ+.

    () For any prime p>5, we have

    (1p)p1k=0TkZk32k{4x22p (mod p2)if p1,4 (mod 15)  p=x2+15y2 (x,yZ),12x22p (mod p2)if p2,8 (mod 15)  p=3x2+5y2 (x,yZ),0 (mod p2)if (15p)=1. (5.28)

    Conjecture 5.3. (ⅰ) For any nZ+, we have

    1nn1k=0(1)k(21k+5)Tk(1,16)Zk252n1kZ+. (5.29)

    () Let p>3 be a prime with p7. Then

    p1k=021k+5(252)kTk(1,16)Zkp3(16(7p)(1p)) (mod p2). (5.30)

    If (7p)=1, then

    1(pn)2(pn1k=021k+5(252)kTk(1,16)Zkp(1p)n1k=021k+5(252)kTk(1,16)Zk)Zp (5.31)

    for all nZ+.

    () For any prime p>3 with p7, we have

    p1k=0Tk(1,16)Zk(252)k{4x22p (mod p2)if p1,2,4 (mod 7)  p=x2+7y2 (x,yZ),0 (mod p2)if p3,5,6 (mod 7). (5.32)

    Conjecture 5.4. (i) For any nZ+, we have

    n | n1k=0kTk(14,1)Zk192n1k. (5.33)

    () Let p>3 be a prime. Then

    p1k=0k192kTk(14,1)Zkp9((1p)(2p)) (mod p2). (5.34)

    If p1,3 (mod 8), then

    1(pn)2(pn1k=0k192kTk(14,1)Zkp(1p)n1k=0k192kTk(14,1))Zp (5.35)

    for all nZ+.

    () For any prime p>3, we have

    (3p)p1k=0Tk(14,1)Zk192k{4x22p (mod p2)if p1,3 (mod 8)  p=x2+2y2 (x,yZ),0 (mod p2)if p5,7 (mod 8). (5.36)

    Conjecture 5.5. () For any nZ+, we have

    n | n1k=0(30k7)Tk(62,1)Zk640n1k. (5.37)

    () Let p be an odd prime with p5. Then

    p1k=030k7640kTk(62,1)Zkp(2(1p)9(15p)) (mod p2). (5.38)

    If (15p)=1, then

    1(pn)2(pn1k=030k7640kTk(62,1)Zkp(1p)n1k=030k7640kTk(62,1)Zk)Zp (5.39)

    for all nZ+.

    () For any prime p>5, we have

    (1p)p1k=0Tk(62,1)Zk640k{4x22p (mod p2)if (2p)=(p3)=(p5)=1  p=x2+30y2,8x22p (mod p2)if (2p)=1, (p3)=(p5)=1  p=2x2+15y2,2p12x2 (mod p2)if (p3)=1, (2p)=(p5)=1  p=3x2+10y2,20x22p (mod p2)if (p5)=1, (2p)=(p3)=1  p=5x2+6y2,0 (mod p2)if (30p)=1, (5.40)

    where x and y are integers.

    Sun [36,37] obtained some supercongruences involving the Franel numbers fn=nk=0(nk)3 (nN). M. Rogers and A. Straub [30] confirmed the 520-series for 1/π involving Franel polynomials conjectured by Sun [34].

    Let p be an odd prime. By [24,Lemma 2.6], we have fk(8)kfp1k (mod p) for each k=0,,p1. Combining this with Remark 1.3(ⅱ), we see that

    p1k=0fkTk(b,c)mk(b24cp)p1k=0(8(b24c)m)kfp1kTp1k(b,c)(b24cp)p1k=0fkTk(b,c)(8(4cb2)/m)k (mod p)

    for any b,c,mZ with p(b24c)m.

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

    k=0(18k+723)fkPk(1+36)(2326)k=27+1132π.

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

    Conjecture 6.1. We have

    k=03k+1(48)kfkTk(4,2)=423π, (6.1)
    k=099k+23(288)kfkTk(8,2)=392π, (6.2)
    k=0105k+17480kfkTk(8,1)=9253π, (6.3)
    k=045k2441kfkTk(47,1)=48354π, (6.4)
    k=0165k+46(2352)kfkTk(194,1)=11253π, (6.5)
    k=042k+511616kfkTk(482,1)=374215π, (6.6)
    k=0990k+3111200kfkTk(898,1)=6807π, (6.7)
    k=0585k+172(13552)kfkTk(1454,1)=1107π, (6.8)
    k=090k+11101568kfkTk(2114,1)=92157π, (6.9)
    k=094185k+17014(105984)kfkTk(2302,1)=852023π, (6.10)
    k=05355k+1381(61952)kfkTk(4354,1)=9687π, (6.11)
    k=0210k+23475904kfkTk(16898,1)=2912231297π. (6.12)

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

    Conjecture 6.2. () For any nZ+, we have

    1nn1k=0(105k+17)480n1kfkTk(8,1)Z+. (6.13)

    () Let p>5 be a prime. Then

    p1k=0105k+17480kfkTk(8,1)p9(161(5p)8) (mod p2). (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] E. S. A. El-Sherpieny, M. A. Ahmed, On the Kumaraswamy Kumaraswamy distribution, Int. J. Basic Appl. Sci., 3 (2014), 372–381. https://doi.org/10.14419/ijbas.v3i4.3182 doi: 10.14419/ijbas.v3i4.3182
    [2] E. Altun, G. G. Hamedani, The log-xgamma distribution with inference and application, J. Soc. Fr. Stat., 159 (2018), 40–55.
    [3] J. Mazucheli, S. R. Bapat, A. F. B. Menezes, A new one-parameter unit-Lindley distribution, Chilean J. Stat., 11 (2020), 53–67.
    [4] J. Mazucheli, A. F. Menezes, S. Dey, Unit-Gompertz distribution with applications, Statistica, 79 (2019), 25–43. https://doi.org/10.6092/issn.1973-2201/8497 doi: 10.6092/issn.1973-2201/8497
    [5] J. Mazucheli, A. F. B. Menezes, L. B. Fernandes, R. P. De Oliveira, M. E. Ghitany, The unit-Weibull distribution as an alternative to the Kumaraswamy distribution for the modeling of quantiles conditional on covariates, J. Appl. Stat., 47 (2020), 954–974. https://doi.org/10.1080/02664763.2019.1657813 doi: 10.1080/02664763.2019.1657813
    [6] K. Modi, V. Gill, Unit Burr-Ⅲ distribution with application, J. Stat. Manag. Syst., 23 (2020), 579–592. https://doi.org/10.1080/09720510.2019.1646503 doi: 10.1080/09720510.2019.1646503
    [7] R. A. R. Bantan, C. Chesneau, J. Farrukh, M. Elgarhy, M. H. Tahir, A. Ali, et al., Some new facts about the unit-Rayleigh distribution with applications, Mathematics, 8 (2020), 1954. https://doi.org/10.3390/math8111954 doi: 10.3390/math8111954
    [8] M. Ç. Korkmaz, C. Chesneau, On the unit Burr-XII distribution with the quantile regression modeling and applications, Comput. Appl. Math., 40 (2021), 29. https://doi.org/10.1007/s40314-021-01418-5 doi: 10.1007/s40314-021-01418-5
    [9] H. S. Bakouch, A. S. Nik, A. Asgharzadeh, H. S. Salinas, A flexible probability model for proportion data: Unit-half-normal distribution, Commun. Stat. Case Stud. Data Anal. Appl., 7 (2021), 271–288. https://doi.org/10.1080/23737484.2021.1882355 doi: 10.1080/23737484.2021.1882355
    [10] M. Irshad, V. Dcruz, R. Maya, The exponentiated unit Lindley distribution: Properties and applications, Ricerche Mat., 73 (2021), 1121–1143. https://doi.org/10.1007/s11587-021-00663-4 doi: 10.1007/s11587-021-00663-4
    [11] A. Krishna, R. Maya, C. Chesneau, M. R. Irshad, The unit Teissier distribution and its applications, Math. Comput. Appl., 27 (2022), 12. http://dx.doi.org/10.3390/mca27010012 doi: 10.3390/mca27010012
    [12] A. I. Al-Omari, A. R. Alanzi, S. S. Alshqaq, The unit two parameters Mirra distribution: Reliability analysis, properties, estimation and applications, Alex. Eng. J., 92 (2024), 238–253. https://doi.org/10.1016/j.aej.2024.02.063 doi: 10.1016/j.aej.2024.02.063
    [13] R. Maya, P. Jodra, M. Irshad, A. Krishna, The unit Muth distribution: statistical properties and applications, Ricerche Mat., 73 (2022), 1843–1866. https://doi.org/10.1007/s11587-022-00703-7 doi: 10.1007/s11587-022-00703-7
    [14] N. Khodja, A. M. Gemeay, H. Zeghdoudi, K. Karakaya, A. M. Alshangiti, M. E. Bakr, et al., Modeling voltage real data set by a new version of Lindley distribution, IEEE Access, 11 (2023), 67220–67229. https://doi.org/10.1109/ACCESS.2023.3287926 doi: 10.1109/ACCESS.2023.3287926
    [15] A. Beghriche, H. Zeghdoudi, V. Raman, S. Chouia, New polynomial exponential distribution: Properties and applications, Statist. Transit., 23 (2022), 95–112. https://doi.org/10.2478/stattrans-2022-0032 doi: 10.2478/stattrans-2022-0032
    [16] A. A. Abd EL-Baset, M. Ghazal, Exponentiated additive Weibull distribution, Reliab. Eng. Syst. Saf., 193 (2020), 106663. http://dx.doi.org/10.1016/j.ress.2019.106663 doi: 10.1016/j.ress.2019.106663
    [17] M. Irshad, D. Shibu, R. Maya, V. Dcruz, Binominal mixture Lindley distribution: Properties and applications, J. Indian Soc. Probab. Stat., 21 (2020), 437–469. https://doi.org/10.1007/s41096-020-00090-y doi: 10.1007/s41096-020-00090-y
    [18] A. Khalil, M. Ijaz, K. Ali, W. K. Mashwani, M. Shafiq, P. Kumam, et al., A novel flexible additive Weibull distribution with real-life applications, Comm. Statist. Theory Methods, 50 (2021), 1557–1572. https://doi.org/10.1080/03610926.2020.1732658 doi: 10.1080/03610926.2020.1732658
    [19] M. E. Ghitany, J. Mazucheli, A. F. B. Menezes, F. Alqallaf, The unit-inverse Gaussian distribution: A new alternative to two-parameter distributions on the unit interval, Commun. Stat. Theory Methods, 48 (2019), 3423–3438. https://doi.org/10.1080/03610926.2018.1476717 doi: 10.1080/03610926.2018.1476717
    [20] M. Ç. Korkmaz, A new heavy-tailed distribution defined on the bounded interval: The logit Slash distribution and its application, J. Appl. Stat., 47 (2020), 2097–2119. https://doi.org/10.1080/02664763.2019.1704701 doi: 10.1080/02664763.2019.1704701
    [21] Y. A. Iriarte, M. de Castro, H. W. Gómez, An alternative one-parameter distribution for bounded data modeling generated from the Lambert transformation, Symmetry, 13 (2021), 1190. http://dx.doi.org/10.3390/sym13071190 doi: 10.3390/sym13071190
    [22] J. Mazucheli, M. Ç. Korkmaz, A. F. B. Menezes, V. Leiva, The unit generalized half-normal quantile regression model: Formulation, estimation, diagnostics, and numerical applications, Soft Comput., 27 (2023), 279–295. http://dx.doi.org/10.1007/s00500-022-07278-3 doi: 10.1007/s00500-022-07278-3
    [23] F. Lad, G. Sanfilippo, G. Agro, Extropy: Complementary dual of entropy, Statist. Sci., 30 (2015), 40–58. http://doi.org/10.1214/14-STS430 doi: 10.1214/14-STS430
    [24] C. Tsallis, Possible generalization of Boltzmann-Gibbs statistics, J. Stat. Phys., 52 (1988), 479–487. http://dx.doi.org/10.1007/BF01016429 doi: 10.1007/BF01016429
    [25] A. Rényi, On measures of entropy and information, Berkeley Symp. Math. Statist. Prob., 4.1 (1961), 547–562.
    [26] J. Havrda, F. Charvat, Quantification method of classification processes, concept of structural \alpha-entropy, Kybernetika, 3 (1967), 30–35.
    [27] A. Al-Shomrani, O. Arif, A. Shawky, S. Hanif, M. Q. Shahbaz, Topp-Leone family of distributions: Some properties and application, Pak. J. Stat. Oper. Res., 12 (2016), 443–451. http://dx.doi.org/10.18187/pjsor.v12i3.1458 doi: 10.18187/pjsor.v12i3.1458
    [28] P. Kumaraswamy, A generalized probability density function for double-bounded random processes, J. Hydrol., 46 (1980), 79–88. https://doi.org/10.1016/0022-1694(80)90036-0 doi: 10.1016/0022-1694(80)90036-0
    [29] A. Pourdarvish, S. M. T. K. Mirmostafaee, K. Naderi, The exponentiated Topp-Leone distribution: Properties and application, J. Appl. Environ. Biol. Sci., 5 (2015), 251–256.
    [30] M. Caramanis, J. Stremel, W. Fleck, S. Daniel, Probabilistic production costing: An investigation of alternative algorithms, Int. J. Elec. Power Energy Syst., 5 (1983), 75–86. https://doi.org/10.1016/0142-0615(83)90011-X doi: 10.1016/0142-0615(83)90011-X
    [31] M. Mazumdar, D. P. Gaver, On the computation of power-generating system reliability indexes, Technometrics, 26 (1984), 173–185. https://doi.org/10.1080/00401706.1984.10487942 doi: 10.1080/00401706.1984.10487942
    [32] P. Sudsila, A. Thongteeraparp, S. Aryuyuen, W. Bodhisuwan, The generalized distributions on the unit interval based on the T-Topp-Leone family of distributions, Trends Sci., 19 (2022), 6186–6186. https://doi.org/10.48048/tis.2022.6186
    [33] M. V. Aarset, How to identify bathtub hazard rate, IEEE Trans. Reliab., 36 (1987), 106–108. https://doi.org/10.1109/TR.1987.5222310 doi: 10.1109/TR.1987.5222310
  • 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
  • © 2024 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(1113) PDF downloads(69) Cited by(1)

Figures and Tables

Figures(9)  /  Tables(7)

/

DownLoad:  Full-Size Img  PowerPoint
Return
Return

Catalog