Processing math: 28%
Research article

Pseudo-3D electrical resistivity tomography imaging of subsurface structure of a sinkhole—A case study in Greene County, Missouri

  • Conventional 2D electrical resistivity tomography (ERT) data were acquired along 16 parallel traverses spaced at 6.1 m (20 ft) intervals across a karst sinkhole site in Greene County Missouri. The acquired ERT data were processed as both 2D data and pseudo-3D data. Based on the correlation with the available core hole control, multichannel analysis of surface waves (MASW) data and field observations, it is concluded that the subsurface structure of the sinkhole is more reliably imaged on the pseudo-3D dataset than in the 2D dataset. The interpretation results of the pseudo-3D ERT indicated that the sinkhole developed at the intersection of three vertical solution-widened joint sets.

    Citation: Shishay Kidanu, Aleksandra Varnavina, Neil Anderson, Evgeniy Torgashov. Pseudo-3D electrical resistivity tomography imaging of subsurface structure of a sinkhole—A case study in Greene County, Missouri[J]. AIMS Geosciences, 2020, 6(1): 54-70. doi: 10.3934/geosci.2020005

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  • Conventional 2D electrical resistivity tomography (ERT) data were acquired along 16 parallel traverses spaced at 6.1 m (20 ft) intervals across a karst sinkhole site in Greene County Missouri. The acquired ERT data were processed as both 2D data and pseudo-3D data. Based on the correlation with the available core hole control, multichannel analysis of surface waves (MASW) data and field observations, it is concluded that the subsurface structure of the sinkhole is more reliably imaged on the pseudo-3D dataset than in the 2D dataset. The interpretation results of the pseudo-3D ERT indicated that the sinkhole developed at the intersection of three vertical solution-widened joint sets.


    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

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

    and

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

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

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

    and

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

    respectively. Therefore

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

    and hence

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

    satisfies the recurrence relation

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

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

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

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

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

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

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

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

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

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

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

    with the aid of the Chu-Vandermonde identity. Therefore

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

    This proves (2.21).

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

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

    Proof. Let n = k+l . Then

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

    and

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

    Hence

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

    This proves (2.22).

    To prove Theorem 1.3, we need an auxiliary theorem.

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

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

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

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

    and similarly

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

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

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

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

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

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

    and

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

    Recall that

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

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

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

    converges. Thus

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

    and hence by the above we have

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

    and

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

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

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

    and

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

    In view of (2.25) and (2.20),

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

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

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

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

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

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

    Therefore

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

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

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

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

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

    Conjecture 3.1. We have the following identities:

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

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

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

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

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

    and (I5) and (I5 ' ) imply that

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

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

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

    and

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

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

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

    and

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

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

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

    and

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

    By [40,Theorem 5.1], we have

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

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

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

    Conjecture 3.2. We have the following identities:

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

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

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

    and that

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

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

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

    Conjecture 3.3. We have the following identities:

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

    and

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

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

    Conjecture 3.4. We have the following identities:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    Conjecture 3.6. We have

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    (ⅱ) Let p>3 be a prime. Then

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

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

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

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

    (ⅲ) Let p>5 be a prime. Then

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

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

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

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

    (ⅱ) Let p>3 be a prime. Then

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

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

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

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

    (ⅲ) Let p>5 be a prime. Then

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

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

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

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

    (ⅱ) Let p>5 be a prime. Then

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

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

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

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

    (ⅲ) Let p>5 be a prime. Then

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

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

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

    diverges.

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

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

    (ⅱ) Let p>3 be a prime. Then

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

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

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

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

    (ⅲ Let p>7 be a prime. Then

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

    where x and y are integers.

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

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

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

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

    (ⅱ) Let p>5 be a prime. Then

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

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

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

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

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

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

    where x and y are integers.

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

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

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

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

    (ⅱ) Let p>3 be a prime. Then

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

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

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

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

    (ⅲ) For any prime p>3 , we have

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

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

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

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

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

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

    and moreover

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

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

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

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

    where x and y are integers.

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

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

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

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

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

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

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

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

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

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

    where x and y are integers.

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

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

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

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

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

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

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

    where x and y are integers.

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

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

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

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

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

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

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

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

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

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

    where x and y are integers.

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

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

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

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

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

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

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

    where x and y are integers.

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

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

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

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

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

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

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

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

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

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

    where x and y are integers.

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

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

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

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

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

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

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

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

    where x and y are integers.

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

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

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

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

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

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

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

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

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

    where x and y are integers.

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

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

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

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

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

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

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

    where x and y are integers.

    Conjecture 4.11. For any odd prime p ,

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

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

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

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

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

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

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

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

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

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

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

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

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

    \rm(ⅱ) Let p>3 be a prime. Then

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    where x and y are integers.

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

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

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

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

    Remark 4.1. We also have some similar conjectures involving

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

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

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

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

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

    and

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

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

    The numbers

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    If ( \frac 7p) = 1 , then

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

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

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

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

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

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

    \rm (ⅱ) Let p>3 be a prime. Then

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    where x and y are integers.

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

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

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

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

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

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

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

    Conjecture 6.1. We have

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

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

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

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

    \rm (ⅱ) Let p>5 be a prime. Then

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    \rm (ⅲ) Let p>5 be a prime. Then

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

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

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

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

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

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

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

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

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

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

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

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

    where x and y are integers.

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

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

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

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

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

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

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

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

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

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

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

    where x and y are integers.

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

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

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

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

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

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

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

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

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

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

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

    where x and y are integers.

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

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

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

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

    For n\in{\Bbb N} let

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

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

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

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

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

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

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

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

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

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

    Conjecture 7.1. We have the following identities.

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

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

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

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

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

    \rm(ⅱ) Let p>7 be a prime. Then

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

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

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

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

    \rm(ⅲ) Let p>7 be a prime. Then

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

    where x and y are integers.

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

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

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

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

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

    \rm(ⅱ) Let p>7 be a prime. Then

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

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

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

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

    \rm(ⅲ) Let p>11 be a prime. Then

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

    where x and y are integers.

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

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

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

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

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

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

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

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

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

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

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

    where x and y are integers.

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

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

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

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

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

    \rm(ⅱ) Let p>7 be a prime. Then

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

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

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

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

    \rm(ⅲ) Let p>11 be a prime. Then

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

    where x and y are integers.

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

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

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

    Theorem 7.1. We have

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

    where

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

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

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

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

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

    Since

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

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

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

    As

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

    by Cooper [9], we finally get

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

    This concludes the proof of (7.29).

    Recall that the numbers

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

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

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

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

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

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

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

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

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

    Conjecture 8.1. (ⅰ) We have

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

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

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

    \rm(ⅱ) Let p>5 be a prime. Then

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

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

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

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

    \rm(ⅲ) Let p>5 be a prime. Then

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

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

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

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

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

    \rm(ⅱ) Let p be a prime. Then

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

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

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

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

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

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

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

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

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

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

    \rm(ⅱ) Let p>3 be a prime. Then

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    \rm(ⅱ) Let p>3 be a prime. Then

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    \rm(ⅲ) Let p>3 be a prime. Then

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

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

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

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

    \rm(ⅱ) Let p>5 be a prime. Then

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

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

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

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

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

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

    The numbers

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

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

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

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

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

    Proof. Note that

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

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

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

    By induction,

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

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

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

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

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

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

    Then

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

    and hence

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

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

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

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

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

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

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

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

    Conjecture 9.1. We have

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

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

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

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

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

    \rm(ⅱ) Let p>3 be a prime. Then

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    Now we give one more conjecture in this section.

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

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

    \rm(ⅱ) Let p>3 be a prime. Then

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

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

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

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

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

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

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

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

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

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

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

    in view of Remark 1.3(ⅱ).

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

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

    implies that

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

    Motivated by this, we pose the following curious conjecture.

    Conjecture 10.1. We have the following identities:

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

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

    Now we state our related conjectures on congruences.

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

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

    and

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

    Conjecture 10.3. (ⅰ) We have

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

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

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

    for each prime p>3 .

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

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

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

    Let p be an odd prime. We conjecture that

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

    and

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

    Though (10.6) implies the congruence

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

    and (10.7) with p>3 implies the congruence

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

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

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

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



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