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New series for powers of π and related congruences

  • Via symbolic computation we deduce 97 new type series for powers of π related to Ramanujan-type series. Here are three typical examples:

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

    with

    P(k)=637379600041024803108k2+657229991696087780968k+19850391655004126179,

    k=1(3k+1)16k(2k+1)2k3(2kk)3=π282,

    and

    n=03n+1(100)nnk=0(nk)2Tk(1,25)Tnk(1,25)=258π,

    where the generalized central trinomial coefficient Tk(b,c) denotes the coefficient of xk in the expansion of (x2+bx+c)k. We also formulate a general characterization of rational Ramanujan-type series for 1/π via congruences, and pose 117 new conjectural series for powers of π via looking for corresponding congruences. For example, we conjecture that

    k=039480k+7321(29700)kTk(14,1)Tk(11,11)2=67955π.

    Eighteen of the new series in this paper involve some imaginary quadratic fields with class number 8.

    Citation: Zhi-Wei Sun. New series for powers of π and related congruences[J]. Electronic Research Archive, 2020, 28(3): 1273-1342. doi: 10.3934/era.2020070

    Related Papers:

    [1] Zhi-Wei Sun . New series for powers of $ \pi $ and related congruences. Electronic Research Archive, 2020, 28(3): 1273-1342. doi: 10.3934/era.2020070
    [2] Harman Kaur, Meenakshi Rana . Congruences for sixth order mock theta functions $ \lambda(q) $ and $ \rho(q) $. Electronic Research Archive, 2021, 29(6): 4257-4268. doi: 10.3934/era.2021084
    [3] Jorge Garcia Villeda . A computable formula for the class number of the imaginary quadratic field $ \mathbb Q(\sqrt{-p}), \ p = 4n-1 $. Electronic Research Archive, 2021, 29(6): 3853-3865. doi: 10.3934/era.2021065
    [4] Fedor Petrov, Zhi-Wei Sun . Proof of some conjectures involving quadratic residues. Electronic Research Archive, 2020, 28(2): 589-597. doi: 10.3934/era.2020031
    [5] Jin-Yun Guo, Cong Xiao, Xiaojian Lu . On $ n $-slice algebras and related algebras. Electronic Research Archive, 2021, 29(4): 2687-2718. doi: 10.3934/era.2021009
    [6] Dušan D. Repovš, Mikhail V. Zaicev . On existence of PI-exponents of unital algebras. Electronic Research Archive, 2020, 28(2): 853-859. doi: 10.3934/era.2020044
    [7] Victor J. W. Guo . A family of $ q $-congruences modulo the square of a cyclotomic polynomial. Electronic Research Archive, 2020, 28(2): 1031-1036. doi: 10.3934/era.2020055
    [8] Chen Wang . Two congruences concerning Apéry numbers conjectured by Z.-W. Sun. Electronic Research Archive, 2020, 28(2): 1063-1075. doi: 10.3934/era.2020058
    [9] Dmitry Krachun, Zhi-Wei Sun . On sums of four pentagonal numbers with coefficients. Electronic Research Archive, 2020, 28(1): 559-566. doi: 10.3934/era.2020029
    [10] Hai-Liang Wu, Zhi-Wei Sun . Some universal quadratic sums over the integers. Electronic Research Archive, 2019, 27(0): 69-87. doi: 10.3934/era.2019010
  • Via symbolic computation we deduce 97 new type series for powers of π related to Ramanujan-type series. Here are three typical examples:

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

    with

    P(k)=637379600041024803108k2+657229991696087780968k+19850391655004126179,

    k=1(3k+1)16k(2k+1)2k3(2kk)3=π282,

    and

    n=03n+1(100)nnk=0(nk)2Tk(1,25)Tnk(1,25)=258π,

    where the generalized central trinomial coefficient Tk(b,c) denotes the coefficient of xk in the expansion of (x2+bx+c)k. We also formulate a general characterization of rational Ramanujan-type series for 1/π via congruences, and pose 117 new conjectural series for powers of π via looking for corresponding congruences. For example, we conjecture that

    k=039480k+7321(29700)kTk(14,1)Tk(11,11)2=67955π.

    Eighteen of the new series in this paper involve some imaginary quadratic fields with class number 8.



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

    k=0bk+cmka(k)=λdπ,()

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

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

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

    p1k=0bk+cmka(k)cp(εddp) (mod p3),

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

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

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

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

    n=05n+164nDn=83π,

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

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

    The author [45,Conjecture 77] conjectured further that

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

    for each odd prime p and positive integer n.

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

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

    Note also that

    T0(b,c)=1,  T1(b,c)=b,

    and

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

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

    For nN the Legendre polynomial of degree n is defined by

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

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

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

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

    Tn(b,c)(b+2c)n+1/224cnπas n+

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

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

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

    Type Ⅰ. k=0a+dkmk(2kk)2Tk(b,c).

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

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

    Type Ⅳ. k=0a+dkmk(2kk)2T2k(b,c).

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

    Type Ⅵ. k=0a+dkmkTk(b,c)3,

    Type Ⅶ. k=0a+dkmk(2kk)Tk(b,c)2,

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

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

    as well as its p-adic analogue

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

    where p is any prime greater than 3.

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

    k=0(4k1)(2kk)4(2k1)4256k=8π2.

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

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

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

    Theorem 1.1. We have the following identities:

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

    and

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

    where

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

    Recall that the Catalan numbers are given by

    Cn:=(2nn)n+1=(2nn)(2nn+1)  (nN).

    For kN it is easy to see that

    (2kk)2k1={1if k=0,2Ck1if k>0.

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

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

    For example, (1.2) has the equivalent form

    k=04k+3(64)kC3k=816π.(1.2)

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

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

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

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

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

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

    and

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

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

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

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

    and

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

    Now we state our second theorem.

    Theorem 1.2. We have the identities

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

    where

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

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

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

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

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

    For example, (1.77) has the following equivalent form

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

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

    Sn:=nk=0(nk)2TkTnk  (n=0,1,2,).

    The values of Sn (n=0,,10) are

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

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

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

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

    Now we state our third theorem.

    Theorem 1.3. We have

    k=07k+324kSk(1,6)=152π, (1.88)
    k=012k+5(28)kSk(1,7)=67π, (1.89)
    k=084k+2980kSk(1,20)=2415π, (1.90)
    k=03k+1(100)kSk(1,25)=258π, (1.91)
    k=0228k+67224kSk(1,56)=807π, (1.92)
    k=0399k+101(676)kSk(1,169)=25358π, (1.93)
    k=02604k+5632600kSk(1,650)=850393π, (1.94)
    k=039468k+7817(6076)kSk(1,1519)=441031π, (1.95)
    k=041667k+78799800kSk(1,2450)=4042564π, (1.96)
    k=074613k+10711(5302)kSk(1,2652)=161517548π. (1.97)

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

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

    Type Ⅷ. k=0a+dkmkTk(b,c)Tk(b,c)2,

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

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

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

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

    k=0bk+cmkak=ri=1λidiπ (1.98)

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

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

    where εi{±1}, εi=1 if di is not an integer square, and c2=c3=0 if r=1 and ε1=1.

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

    Conjecture 1.2. Let (an)n0 be an integer sequence with no a,xZ such that an=a(2nn)nk=0(2kk)2(2(nk)nk)xnk for all nN, and let b,c,m,d1,d2,d3Z with bcm0. Assume that limn+n|an|=r<|m|, and πk=0bk+cmkak is an algebraic number. Suppose that c1,c2,c3Q with c1+c2+c3=a0c , and

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

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

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

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

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

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

    p1k=0(8k+5)T2k3p(3p) (mod p2)

    for any prime p>3, which was confirmed by Y.-P. Mu and Z.-W. Sun [26]. This is not a counterexample to Conjecture 1.3 since k=0(8k+5)T2k diverges.

    All the new series and related congruences in Sections 3-9 support Conjectures 1.1-1.3. We discover the conjectural series for 1/π in Sections 3-9 based on the author's previous PhilosophyaboutSeriesfor 1/π} stated in [35], the PSLQ algorithm to discover integer relations (cf. [13]), and the following DualityPrinciple based on the author's experience and intuition.

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

    ak(dp)Dkap1k (mod p) (1.101)

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

    k=0bk+cmkak=λ1d1+λ2d2+λ3d3π

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

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

    for any prime p>3 with pdD.

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

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

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

    for all k=0,1,,p1.

    For a series k=0ak with a0,a1, real numbers, if limk+ak+1/ak=r(1,1) then we say that the series converge at a geometric rate with ratio r. Except for (7.1), all other conjectural series in Sections 3-9 converge at geometric rates and thus one can easily check them numerically via a computer.

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

    Lemma 2.1. Let m0 and n0 be integers. Then

    nk=0((64m)k332k216k+8)(2kk)3(2k1)2mk=8(2n+1)mn(2nn)3, (2.1)
    nk=0((64m)k396k2+48k8)(2kk)3(2k1)3mk=8mn(2nn)3, (2.2)
    nk=0((108m)k354k212k+6)(2kk)2(3kk)(2k1)(3k1)mk=6(3n+1)mn(2nn)2(3nn), (2.3)
    nk=0((108m)k3(54+m)k212k+6)(2kk)2(3kk)(k+1)(2k1)(3k1)mk=6(3n+1)(n+1)mn(2nn)2(3nn), (2.4)
    nk=0((256m)k3128k216k+8)(2kk)2(4k2k)(2k1)(4k1)mk=8(4n+1)mn(2nn)2(4n2n), (2.5)
    nk=0((256m)k3(128+m)k216k+8)(2kk)2(4k2k)(k+1)(2k1)(4k1)mk=8(4n+1)(n+1)mn(2nn)2(4n2n), (2.6)
    nk=0((1728m)k3864k248k+24)(2kk)(3kk)(6k3k)(2k1)(6k1)mk=24(6n+1)mn(2nn)(3nn)(6n3n), (2.7)
    nk=0((1728m)k3(864+m)k248k+24)(2kk)(3kk)(6k3k)(k+1)(2k1)(6k1)mk=24(6n+1)(n+1)mn(2nn)(3nn)(6n3n). (2.8)

    Remark 2.1. The eight identities in Lemma 2.1 can be easily proved by induction on n. In light of Stirling's formula, n!2πn(n/e)n as n+, we have

    (2nn)4nnπ,  (2nn)(3nn)327n2nπ, (2.9)
    (2nn)(4n2n)64n2nπ,  (3nn)(6nn)432n2nπ. (2.10)

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

    By (2.1) with m=64, we have

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

    Note that

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

    and recall Bauer's series

    k=0(4k+1)(2kk)3(64)k=2π.

    So, we get

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

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

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

    and hence

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

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

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

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

    and hence

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

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

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

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

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

    and hence

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

    This proves (1.71).

    By (2.8) with m=6403203, we have

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

    and hence

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

    Note that

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

    Therefore, with the help of (1.71) we get

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

    This proves (1.72).

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

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

    nk=1mk((m64)k332k2+16k+8)(2k+1)2k3(2kk)3=mn+1(2n+1)2(2nn)3m, (2.11)
    nk=1mk((m64)k396k248k8)(2k+1)3k3(2kk)3=mn+1(2n+1)3(2nn)3m, (2.12)
    nk=1mk((m108)k354k2+12k+6)(2k+1)(3k+1)k3(2kk)2(3kk)=mn+1(2n+1)(3n+1)(2nn)2(3nn)m, (2.13)
    1<knmk((m108)k3(54+m)k2+12k+6)(k1)(2k+1)(3k+1)k3(2kk)2(3kk)=mn+1n(2n+1)(3n+1)(2nn)2(3nn)m2144, (2.14)
    nk=1mk((m256)k3128k2+16k+8)(2k+1)(4k+1)k3(2kk)2(4k2k)=mn+1(2n+1)(4n+1)(2nn)2(4n2n)m, (2.15)
    1<knmk((m256)k3(128+m)k2+16k+8)(k1)(2k+1)(4k+1)k3(2kk)2(4k2k)=mn+1n(2n+1)(4n+1)(2nn)2(4n2n)m2360. (2.16)

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

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

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

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

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

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

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

    for all nZ+, and hence

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

    Notice that

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

    So we have

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

    and hence (1.77) holds.

    By (2.11) with m=64, we have

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

    for all nZ+, and hence

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

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

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

    we see that

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

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

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

    for all nZ+, and hence

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

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

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

    This proves (1.79).

    By (2.13) with m=27, we have

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

    As

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

    and

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

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

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

    and hence

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

    As

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

    with the aid of (1.82) we get

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

    and hence (1.83) follows.

    Other identities in Theorem 1.2 can be proved similarly.

    For integers nk0, we define

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

    For nN we set

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

    Lemma 2.3. For any nN, we have

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

    and

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

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

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

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

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

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

    with 0ik, we find that

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

    where

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

    and

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

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

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

    and

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

    respectively. Therefore

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

    and hence

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

    satisfies the recurrence relation

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

    As pointed out by J. Franel [14], the Franel numbers satisfy the same recurrence. Note also that u(0)=f0=1 and u(1)=f1=2. So we always have u(n)=fn. This proves (2.19).

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

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

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

    Lemma 2.4. For any cZ and nN, we have

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

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

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

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

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

    with the aid of the Chu-Vandermonde identity. Therefore

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

    This proves (2.21).

    Lemma 2.5. For kN and lZ+, we have

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

    Proof. Let n=k+l. Then

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

    and

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

    Hence

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

    This proves (2.22).

    To prove Theorem 1.3, we need an auxiliary theorem.

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

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

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

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

    and similarly

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

    where we consider (x1) as 0.

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

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

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

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

    and

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

    Recall that

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

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

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

    converges. Thus

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

    and hence by the above we have

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

    and

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

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

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

    and

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

    In view of (2.25) and (2.20),

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

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

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

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

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

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

    Therefore

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

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

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

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

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

    Conjecture 3.1. We have the following identities:

    k=050k+1(256)k(2kk)(2kk+1)Tk(1,16)=83π,(1)
    k=0(100k24k7)(2kk)2Tk(1,16)(2k1)2(256)k=24π,(1

    Remark 3.1. For each , we have

    since and . Thus, for example, [40,(I1)] and (I1) together imply that

    and (I5) and (I5) imply that

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

    and

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

    and

    for each , and that for any prime with we have

    and

    By [40,Theorem 5.1], we have

    for any prime . 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:

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

    and that

    where and 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:

    and

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

    Conjecture 3.4. We have the following identities:

    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 be an odd prime and let . If , then

    divided by is a -adic integer. If , then

    divided by is a -adic integer. If but , then

    divided by is a -adic integer. If but , then

    divided by is a -adic integer. If but , then

    divided by is a -adic integer.

    Now we pose four conjectural series for of type Ⅷ.

    Conjecture 3.6. We have

    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 , we have

    (3.1)

    and this number is odd if and only if is a power of two i.e., .

    (ⅱ) Let be a prime. Then

    (3.2)

    If , then

    (3.3)

    for all .

    (ⅲ) Let be a prime. Then

    (3.4)

    Remark 3.4. The imaginary quadratic field has class number two.

    Conjecture 3.8. (ⅰ) For any , we have

    (3.5)

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

    (ⅱ) Let be a prime. Then

    (3.6)

    If , then

    (3.7)

    for all .

    (ⅲ) Let be a prime. Then

    (3.8)

    Remark 3.5. This conjecture can be viewed as the dual of Conjecture 3.7. Note that the series diverges.

    Conjecture 3.9. (ⅰ) For each , we have

    (3.9)

    (ⅱ) Let be a prime. Then

    (3.10)

    If , then

    (3.11)

    divided by is a -adic integer for each .

    (ⅲ) Let be a prime. Then

    (3.12)

    Remark 3.6. The imaginary quadratic field has class number two.

    Conjecture 3.10. (ⅰ) For each , we have

    (3.13)

    (ⅱ) Let be a prime. Then

    (3.14)

    If , then

    (3.15)

    divided by is a -adic integer for each .

    (ⅲ) Let be a prime. Then

    (3.16)

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

    diverges.

    Conjecture 3.11. (ⅰ) For each , we have

    (3.17)

    (ⅱ) Let be a prime. Then

    (3.18)

    If , then

    (3.19)

    divided by is an -adic integer for any .

    (ⅲ Let be a prime. Then

    (3.20)

    where and are integers.

    Remark 3.8. Note that the imaginary quadratic field has class number .

    Conjecture 3.12. (ⅰ) For each , we have

    (3.21)

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

    (ⅱ) Let be a prime. Then

    (3.22)

    If , then

    (3.23)

    divided by is a -adic integer for each .

    (ⅲ) Let be a prime with . Then

    (3.24)

    where and are integers.

    Remark 3.9. Note that the imaginary quadratic field has class number .

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

    Conjecture 4.1. (ⅰ) For any , we have

    (4.1)

    (ⅱ) Let be a prime. Then

    (4.2)

    If , then

    (4.3)

    for all .

    (ⅲ) For any prime , we have

    (4.4)

    Conjecture 4.2. (ⅰ) For any , we have

    (4.5)

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

    (ⅱ) Let be an odd prime. Then

    (4.6)

    and moreover

    (4.7)

    for all .

    (ⅲ) For any prime , we have

    (4.8)

    where and are integers.

    Conjecture 4.3. (ⅰ) For any , we have

    (4.9)

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

    (ⅱ) Let be an odd prime with . Then

    (4.10)

    If , then

    (4.11)

    for all .

    (ⅲ) For any prime , we have

    (4.12)

    where and are integers.

    Conjecture 4.4. (ⅰ) For any , we have

    (4.13)

    (ⅱ) Let be an odd prime. Then

    (4.14)

    for all .

    (ⅲ) For any prime with , we have

    (4.15)

    where and are integers.

    Conjecture 4.5. (ⅰ) For any , we have

    (4.16)

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

    (ⅱ) Let be an odd prime with . Then

    (4.17)

    If , then

    (4.18)

    for all .

    (ⅲ) For any prime , we have

    (4.19)

    where and are integers.

    Conjecture 4.6. (ⅰ) For any , we have

    (4.20)

    (ⅱ) Let be an odd prime. Then

    (4.21)

    divided by is a -adic integer for any .

    (ⅲ) For any prime with , we have

    (4.22)

    where and are integers.

    Conjecture 4.7. (ⅰ) For any , we have

    (4.23)

    and this number is odd if and only if .

    (ⅱ) Let be an odd prime with . Then

    (4.24)

    If , then

    (4.25)

    divided by is a -adic integer for any .

    (ⅲ)For any odd prime , we have

    (4.26)

    where and are integers.

    Conjecture 4.8. (ⅰ) For any , we have

    (4.27)

    and this number is odd if and only if .

    (ⅱ) Let be an odd prime. Then

    (4.28)

    divided by is a -adic integer for any .

    (ⅲ) For any prime with , we have

    (4.29)

    where and are integers.

    Conjecture 4.9. (ⅰ) For any , we have

    (4.30)

    (ⅱ) Let be an odd prime. Then

    (4.31)

    If , then

    (4.32)

    divided by is a -adic integer for any .

    (ⅲ) For any prime with , we have

    (4.33)

    where and are integers.

    Conjecture 4.10. (ⅰ) For any , we have

    (4.34)

    Let be an odd prime. Then

    (4.35)

    divided by is a -adic integer for any .

    For any prime with , we have

    (4.36)

    where and are integers.

    Conjecture 4.11. For any odd prime ,

    (4.37)

    Also, for any prime we have

    (4.38)

    Conjecture 4.12. For any , we have

    (4.39)

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

    For any odd prime and positive integer , we have

    (4.40)

    Let be an odd prime. Then

    (4.41)

    Conjecture 4.13. For any , we have

    (4.42)

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

    Let be a prime. Then

    (4.43)

    If , then

    (4.44)

    for all .

    For any prime , we have

    (4.45)

    Conjecture 4.14. For any , we have

    (4.46)

    Let be an odd prime. Then

    (4.47)

    If , then

    (4.48)

    for all .

    For any prime , we have

    (4.49)

    where and are integers.

    Conjecture 4.15. Let be an odd prime with . Then

    (4.50)

    where and are integers. If , then

    Remark 4.1. We also have some similar conjectures involving

    modulo , where is a prime greater than .

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

    Conjecture 4.16. For any odd prime and integer , we have

    (4.51)

    and

    (4.52)

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

    The numbers

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

    Let be an odd prime. For any , we have

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

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

    for any with .

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

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

    (5.1)
    (5.2)
    (5.3)
    (5.4)
    (5.5)
    (5.6)
    (5.7)
    (5.8)
    (5.9)
    (5.10)
    (5.11)
    (5.12)
    (5.13)
    (5.14)
    (5.15)
    (5.16)
    (5.17)
    (5.18)
    (5.19)
    (5.20)
    (5.21)
    (5.22)
    (5.23)
    (5.24)

    Below we present some conjectures on congruences related to , , and .

    Conjecture 5.2. (ⅰ) For any , we have

    (5.25)

    Let be an odd prime with . Then

    (5.26)

    If , then

    (5.27)

    for all .

    For any prime , we have

    (5.28)

    Conjecture 5.3. (ⅰ) For any , we have

    (5.29)

    Let be a prime with . Then

    (5.30)

    If , then

    (5.31)

    for all .

    For any prime with , we have

    (5.32)

    Conjecture 5.4. For any , we have

    (5.33)

    Let be a prime. Then

    (5.34)

    If , then

    (5.35)

    for all .

    For any prime , we have

    (5.36)

    Conjecture 5.5. For any , we have

    (5.37)

    Let be an odd prime with . Then

    (5.38)

    If , then

    (5.39)

    for all .

    For any prime , we have

    (5.40)

    where and are integers.

    Sun [36,37] obtained some supercongruences involving the Franel numbers . M. Rogers and A. Straub [30] confirmed the -series for involving Franel polynomials conjectured by Sun [34].

    Let be an odd prime. By [24,Lemma 2.6], we have for each . Combining this with Remark 1.3(ⅱ), we see that

    for any with .

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

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

    Conjecture 6.1. We have

    (6.1)
    (6.2)
    (6.3)
    (6.4)
    (6.5)
    (6.6)
    (6.7)
    (6.8)
    (6.9)
    (6.10)
    (6.11)
    (6.12)

    We now present a conjecture on congruence related to .

    Conjecture 6.2. For any , we have

    (6.13)

    Let be a prime. Then

    (6.14)

    If , then

    (6.15)

    for all .

    For any prime , we have

    (6.16)

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

    Conjecture 6.3. For any , we have

    (6.17)

    Let be an odd prime. Then

    (6.18)

    If , then

    (6.19)

    divided by is a -adic integer for any .

    Let be a prime. Then

    (6.20)

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

    The following conjecture is related to the identity .

    Conjecture 6.4. For any , we have

    (6.21)

    Let be a prime with . Then

    (6.22)

    If , then

    (6.23)

    divided by is a -adic integer for any .

    Let be a prime with . Then

    (6.24)

    where and are integers.

    Remark 6.3. Note that the imaginary quadratic field has class number .

    The following conjecture is related to the identity .

    Conjecture 6.5. For any , we have

    (6.25)

    Let be a prime with . Then

    (6.26)

    If , then

    (6.27)

    divided by is a -adic integer for any .

    Let be a prime with . Then

    (6.28)

    where and are integers.

    Remark 6.4. Note that the imaginary quadratic field has class number .

    The following conjecture is related to the identity .

    Conjecture 6.6. For any , we have

    (6.29)

    Let be an odd prime with . Then

    (6.30)

    If , then

    (6.31)

    divided by is a -adic integer for any .

    Let be a prime with . Then

    (6.32)

    where and are integers.

    Remark 6.5. Note that the imaginary quadratic field has class number .

    The identities are related to the quadratic fields

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

    For let

    It is known that for all . See [43,20,26] for some congruences on polynomials related to these numbers.

    Let be a prime. For any , we have

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

    for any with .

    Wan and Zudilin [49] obtained the following irrational series for involving the Legendre polynomials and the sequence :

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

    Conjecture 7.1. We have the following identities.

    (7.1)
    (7.2)
    (7.3)
    (7.4)
    (7.5)
    (7.6)
    (7.7)
    (7.8)
    (7.9)
    (7.10)
    (7.11)
    (7.12)

    Now we present a conjecture on congruences related to .

    Conjecture 7.2. For any , we have

    (7.13)

    and this number is odd if and only if .

    Let be a prime. Then

    (7.14)

    If , then

    (7.15)

    divided by is a -adic integer for any .

    Let be a prime. Then

    (7.16)

    where and are integers.

    Remark 7.1. Note that the imaginary quadratic field has class number .

    The following conjecture is related to the identity .

    Conjecture 7.3. For any , we have

    (7.17)

    and this number is odd if and only if .

    Let be a prime. Then

    (7.18)

    If , then

    (7.19)

    divided by is a -adic integer for any .

    Let be a prime. Then

    (7.20)

    where and are integers.

    Remark 7.2. Note that the imaginary quadratic field has class number .

    Now we pose a conjecture related to the identity .

    Conjecture 7.4. For any , we have

    (7.21)

    Let be a prime with . Then

    (7.22)

    If , then

    (7.23)

    is a -adic integer for any .

    Let be a prime with . Then

    (7.24)

    where and are integers.

    Remark 7.3. Note that the imaginary quadratic field has class number .

    Now we pose a conjecture related to the identity .

    Conjecture 7.5. For any , we have

    (7.25)

    and this number is odd if and only if .

    Let be a prime. Then

    (7.26)

    If , then

    (7.27)

    divided by is a -adic integer for each .

    Let be a prime. Then

    (7.28)

    where and are integers.

    Remark 7.4. Note that the imaginary quadratic field has class number . We believe that is the largest positive squarefree number for which the imaginary quadratic field can be used to construct a Ramanujan-type series for .

    The identities are related to the imaginary quadratic fields , , , (with class number ) 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 conjectured by the author (cf. [34,(3.28)] and [35,Conjecture 7.9]) in 2011.

    Theorem 7.1. We have

    (7.29)

    where

    Proof. The Franel numbers of order are given by . Note that

    By [11,(8.1)], for and , we have

    (7.30)

    Since

    putting , and in (7.30) we obtain

    As

    by Cooper [9], we finally get

    This concludes the proof of (7.29).

    Recall that the numbers

    are a kind of Apéry numbers. Let be an odd prime. For any , we have

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

    for any with .

    Wan and Zudilin [49] obtained the following irrational series for involving the Legendre polynomials and the numbers :

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

    Conjecture 8.1. (ⅰ) We have

    (8.1)

    Also, for any we have

    (8.2)

    Let be a prime. Then

    (8.3)

    If , then

    (8.4)

    for all .

    Let be a prime. Then

    (8.5)

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

    Conjecture 8.2. For any , we have

    (8.6)

    and this number is odd if and only if .

    Let be a prime. Then

    (8.7)

    If i.e., , then

    (8.8)

    divided by is a -adic integer for any .

    For any prime , we have

    (8.9)

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

    Conjecture 8.3. For any , we have

    (8.10)

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

    Let be a prime. Then

    (8.11)

    If , then

    (8.12)

    for all .

    For any odd prime , we have

    (8.13)

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

    Conjecture 8.4. For any , we have

    (8.14)

    and this number is odd if and only if .

    Let be any odd prime. Then

    (8.15)

    If , then

    (8.16)

    for all .

    For any odd prime , we have

    (8.17)

    Conjecture 8.5. For any , we have

    (8.18)

    Let be a prime. Then

    (8.19)

    If , then

    (8.20)

    for all .

    Let be a prime. Then

    (8.21)

    Conjecture 8.6. For any , we have

    (8.22)

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

    Let be an odd prime. Then

    (8.23)

    If , then

    (8.24)

    for all .

    Let be a prime. Then

    (8.25)

    Conjecture 8.7. For any , we have

    (8.26)

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

    Let be a prime. Then

    (8.27)

    If , then

    (8.28)

    for all .

    For any prime , we have

    (8.29)

    The numbers

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

    Lemma 9.1. Let be a prime. Then

    Proof. Note that

    with the help of the known congruence conjectured by F. Rodriguez-Villegas [28] and proved by E. Mortenson [25]. Similarly,

    By induction,

    for all . In particular,

    So we have for . (Note that and .)

    Now let and assume that

    Then

    and hence

    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 be a prime. In view of Lemma 9.1 and Remark 1.3(ⅱ), we have

    for any with .

    Wan and Zudilin [49] obtained the following irrational series for involving the Legendre polynomials and the numbers :

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

    Conjecture 9.1. We have

    (9.1)
    (9.2)
    (9.3)
    (9.4)
    (9.5)

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

    Conjecture 9.2. For any , we have

    (9.6)

    and this number is odd if and only if .

    Let be a prime. Then

    (9.7)

    If i.e., , then

    (9.8)

    for all .

    For any prime , we have

    (9.9)

    Conjecture 9.3. For any , we have

    (9.10)

    and this number is odd if and only if .

    Let be a prime with . Then

    (9.11)

    If i.e., , then

    (9.12)

    for all .

    For any prime with , we have

    (9.13)

    Now we give one more conjecture in this section.

    Conjecture 9.4. For any integer , we have

    (9.14)

    Let be a prime. Then

    (9.15)

    If , then

    (9.16)

    for all .

    For any prime , we have

    (9.17)

    Remark 9.1. For primes with , in general the congruence (9.16) is not always valid for all . This does not violate Conjecture 1.2 since . If the series converges, its value times should be a rational number.

    Let be an odd prime and let with . Then

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

    in view of Remark 1.3(ⅱ).

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

    implies that

    Motivated by this, we pose the following curious conjecture.

    Conjecture 10.1. We have the following identities:

    (10.1)
    (10.2)

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

    Now we state our related conjectures on congruences.

    Conjecture 10.2. For any prime , we have

    (10.3)

    and

    (10.4)

    Conjecture 10.3. (ⅰ) We have

    for all , and also

    for each prime .

    For any prime and , we have

    (10.5)

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

    Let be an odd prime. We conjecture that

    (10.6)

    and

    (10.7)

    Though (10.6) implies the congruence

    and (10.7) with implies the congruence

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

    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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