Learning deep forest for face anti-spoofing: An alternative to the neural network against adversarial attacks

Rizhao Cai; Liepiao Zhang; Changsheng Chen; Yongjian Hu; Alex Kot; Rizhao Cai; Liepiao Zhang; Changsheng Chen; Yongjian Hu; Alex Kot

doi:10.3934/era.2024259

Electronic Research Archive

2024, Volume 32, Issue 10: 5592-5614. doi: 10.3934/era.2024259

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Research article

Learning deep forest for face anti-spoofing: An alternative to the neural network against adversarial attacks

1.
School of Electrical and Electronic Engineering, Nanyang Technological University, 639798, Singapore
2.
GRGTally-vision I.T. Co., Ltd., Guangzhou 510663, China
3.
College of Electronics and Information Engineering, Shenzhen University, Shenzhen 518061, China
4.
School of Electronic and Information Engineering, South China University of Technology, Guangzhou 510641, China

Received: 20 May 2024 Revised: 31 August 2024 Accepted: 09 September 2024 Published: 11 October 2024

Face anti-spoofing (FAS) is significant for the security of face recognition systems. neural networks (NNs), including convolutional neural network (CNN) and vision transformer (ViT), have been dominating the field of the FAS. However, NN-based methods are vulnerable to adversarial attacks. Attackers could insert adversarial noise into spoofing examples to circumvent an NN-based face-liveness detector. Our experiments show that the CNN or ViT models could have at least an 8% equal error rate (EER) increment when encountering adversarial examples. Thus, developing methods other than NNs is worth exploring to improve security at the system level. In this paper, we have proposed a novel solution for FAS against adversarial attacks, leveraging a deep forest model. Our approach introduces a multi-scale texture representation based on local binary patterns (LBP) as the model input, replacing the grained-scanning mechanism (GSM) used in the traditional deep forest model. Unlike GSM, which scans raw pixels and lacks discriminative power, our LBP-based scheme is specifically designed to capture texture features relevant to spoofing detection. Additionally, transforming the input from the RGB space to the LBP space enhances robustness against adversarial noise. Our method achieved competitive results. When testing with adversarial examples, the increment of EER was less than 3%, more robust than CNN and ViT. On the benchmark database IDIAP REPLAY-ATTACK, a 0% EER was achieved. This work provides a competitive option in a fusing scheme for improving system-level security and offers important ideas to those who want to explore methods besides CNNs. To the best of our knowledge, this is the first attempt at exploiting the deep forest model in the problem of FAS, with the consideration of adversarial attacks.

Keywords:

Citation: Rizhao Cai, Liepiao Zhang, Changsheng Chen, Yongjian Hu, Alex Kot. Learning deep forest for face anti-spoofing: An alternative to the neural network against adversarial attacks[J]. Electronic Research Archive, 2024, 32(10): 5592-5614. doi: 10.3934/era.2024259

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Abstract

1. Introduction and our main results

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

$\sum\limits_{k=0}^{\infty} \frac{b k+c}{m^{k}} a(k)=\frac{\lambda \sqrt{d}}{\pi},\;\;\;\;(*)$

where $b,c,m$ are integers with $bm\not = 0$ , $d$ is a positive squarefree number, $\lambda$ is a nonzero rational number, and $a(k)$ is one of the products

$\binom{2k}k^3,\ \binom{2k}k^2 \binom{3k}k,\ \binom{2k}k^2 \binom{4k}{2k},\ \binom{2k}k \binom{3k}k \binom{6k}{3k}.$

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

$\sum\limits_{k = 0}^{p-1} \frac{bk+c}{m^k}a(k){\equiv} cp \left( \frac{{\varepsilon}_d d}p \right)\ ({\rm{mod}}\ {p^3}),$

where $p$ is any odd prime with $p\nmid dm$ and $\lambda\in{\Bbb Z}_p$ , ${\varepsilon}_1\in\{\pm1\}$ and ${\varepsilon}_d = -1$ if $d>1$ . (As usual, ${\Bbb Z}_p$ denotes the ring of all $p$ -adic integers, and $( \frac{\cdot}p)$ 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

$\frac{\sum_{k = 0}^{pn-1}(21k+8) \binom{2k}k^3-p\sum_{k = 0}^{n-1}(21k+8) \binom{2k}k^3}{(pn)^3 \binom{2n}n^3} \in{\Bbb Z}_p,$

where $p$ is an odd prime, and $n\in{\Bbb Z}^+ = \{1,2,3,\ldots\}$ . 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

$\sum\limits_{n = 0}^\infty \frac{5n+1}{64^n}D_n = \frac 8{\sqrt3\,\pi},$

where $D_n$ denotes the Domb number $\sum_{k = 0}^n \binom nk^2 \binom{2k}k \binom{2(n-k)}{n-k}$ ; Zudilin [53] conjectured its $p$ -adic analogue:

$\sum\limits_{k = 0}^{p-1} \frac{5k+1}{64^k}D_k{\equiv} p \left( \frac p3 \right)\ ({\rm{mod}}\ {p^3})\quad \text{for any prime}\ p > 3.$

The author [45,Conjecture 77] conjectured further that

$\frac1{(pn)^3}\left(\sum\limits_{k = 0}^{pn-1} \frac{5k+1}{64^k}D_k- \left( \frac p3 \right)p\sum\limits_{k = 0}^{n-1} \frac{5k+1}{64^r}D_k\right)\in{\Bbb Z}_p$

for each odd prime $p$ and positive integer $n$ .

Let $b,c\in{\Bbb Z}$ . For each $n\in{\Bbb N} = \{0,1,2,\ldots\}$ , we denote the coefficient of $x^n$ in the expansion of $(x^2+bx+c)^n$ by $T_n(b,c)$ , and call it a generalized central trinomial coefficient. In view of the multinomial theorm, we have

$T_n(b,c) = \sum\limits_{k = 0}^{\lfloor n/2\rfloor} \binom n{2k} \binom{2k}kb^{n-2k}c^k = \sum\limits_{k = 0}^{\lfloor n/2\rfloor} \binom nk \binom{n-k}k b^{n-2k}c^k.$

Note also that

$T_0(b,c) = 1,\ \ T_1(b,c) = b,$

and

$(n+1)T_{n+1}(b,c) = (2n+1)bT_n(b,c)-n(b^2-4c)T_{n-1}(b,c)$

for all $n\in{\Bbb Z}^+.$ Clearly, $T_n(2,1) = \binom{2n}n$ for all $n\in{\Bbb N}$ . Those $T_n: = T_n(1,1)$ with $n\in{\Bbb N}$ are the usual central trinomial coefficients, and they play important roles in enumerative combinatorics. We view $T_n(b,c)$ as a natural generalization of central binomial and central trinomial coefficients.

For $n\in{\Bbb N}$ the Legendre polynomial of degree $n$ is defined by

$P_n(x): = \sum\limits_{k = 0}^n \binom nk \binom{n+k}k \left( \frac{x-1}2 \right)^k.$

It is well-known that if $b,c\in{\Bbb Z}$ and $b^2-4c\not = 0$ then

$T_n(b,c) = (\sqrt{b^2-4c})^nP_n \left( \frac{b}{\sqrt{b^2-4c}} \right)\quad \text{for all}\ n\in{\Bbb N}.$

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

$T_n(b,c)\sim \frac{(b+2\sqrt{c})^{n+1/2}}{2\root{4}\of c\sqrt{n\pi}}\quad \text{as}\ n\to+\infty$

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

$\lim\limits_{n\to\infty}\root n\of{|T_n(b,c)|} = \sqrt{b^2-4c}.$

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

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

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

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

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

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

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

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

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

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

as well as its $p$ -adic analogue

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

where $p$ is any prime greater than $3$ .

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

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

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

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

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

Theorem 1.1. We have the following identities:

$\begin{align} \sum\limits_{k = 0}^\infty \frac{k(4k-1) \binom{2k}k^3}{(2k-1)^2(-64)^k}& = - \frac1{\pi}, \end{align}$

(1.1)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(4k-1) \binom{2k}k^3}{(2k-1)^3(-64)^k}& = \frac2{\pi}, \end{align}$

(1.2)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(12k^2-1) \binom{2k}k^3}{(2k-1)^2 256^k}& = - \frac2{\pi}, \end{align}$

(1.3)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{k(6k-1) \binom{2k}k^3}{(2k-1)^3256^k}& = \frac1{2\pi}, \end{align}$

(1.4)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(28k^2-4k-1) \binom{2k}k^3}{(2k-1)^2(-512)^k}& = - \frac{3\sqrt2}{\pi}, \end{align}$

(1.5)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(30k^2+3k-2) \binom{2k}k^3}{(2k-1)^3(-512)^k}& = \frac{27\sqrt2}{8\pi}, \end{align}$

(1.6)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(28k^2-4k-1) \binom{2k}k^3}{(2k-1)^2 4096^k}& = - \frac3{\pi}, \end{align}$

(1.7)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(42k^2-3k-1) \binom{2k}k^3}{(2k-1)^3 4096^k}& = \frac{27}{8\pi}, \end{align}$

(1.8)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(34k^2-3k-1) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)(-192)^k}& = - \frac{10}{\sqrt3\,\pi}, \end{align}$

(1.9)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(64k^2-11k-7) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)(-192)^k}& = - \frac{125\sqrt3}{9\pi}, \end{align}$

(1.10)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(14k^2+k-1) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)216^k}& = - \frac{\sqrt3}{\pi}, \end{align}$

(1.11)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(90k^2+7k+1) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)216^k}& = \frac{9\sqrt3}{2\pi}, \end{align}$

(1.12)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(34k^2-3k-1) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)(-12)^{3k}}& = - \frac{2\sqrt3}{\pi}, \end{align}$

(1.13)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(17k+5) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)(-12)^{3k}}& = \frac{9\sqrt3}{\pi}, \end{align}$

(1.14)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(111k^2-7k-4) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)1458^k}& = - \frac{45}{4\pi}, \end{align}$

(1.15)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(1524k^2+899k+263) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)1458^k}& = \frac{3375}{4\pi}, \end{align}$

(1.16)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(522k^2-55k-13) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)(-8640)^k}& = - \frac{54\sqrt{15}}{5\pi}, \end{align}$

(1.17)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(1836k^2+2725k+541) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)(-8640)^k}& = \frac{2187\sqrt{15}}{5\pi}, \end{align}$

(1.18)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(529k^2-45k-16) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)15^{3k}}& = - \frac{55\sqrt3}{2\pi}, \end{align}$

(1.19)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(77571k^2+68545k+16366) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)15^{3k}}& = \frac{59895\sqrt3}{2\pi}, \end{align}$

(1.20)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(574k^2-73k-11) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)(-48)^{3k}}& = -20 \frac{\sqrt3}{\pi}, \end{align}$

(1.21)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(8118k^2+9443k+1241) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)(-48)^{3k}}& = \frac{2250\sqrt3}{\pi}, \end{align}$

(1.22)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(978k^2-131k-17) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)(-326592)^k}& = - \frac{990\sqrt7}{49\pi}, \end{align}$

(1.23)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(592212k^2+671387k^2+77219) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)(-326592)^k}& = \frac{4492125\sqrt7}{49\pi}, \end{align}$

(1.24)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(116234k^2-17695k-1461) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)(-300)^{3k}}& = -2650 \frac{\sqrt3}{\pi}, \end{align}$

(1.25)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(223664832k^2+242140765k+18468097) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)(-300)^{3k}}& = 33497325 \frac{\sqrt3}{\pi}, \end{align}$

(1.26)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(122k^2+3k-5) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)648^k}& = - \frac{21}{2\pi}, \end{align}$

(1.27)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(1903k^2+114k+41) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)648^k}& = \frac{343}{2\pi}, \end{align}$

(1.28)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(40k^2-2k-1) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)(-1024)^k}& = - \frac4{\pi}, \end{align}$

(1.29)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(8k^2-2k-1) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)(-1024)^k}& = - \frac{16}{5\pi}, \end{align}$

(1.30)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(176k^2-6k-5) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)48^{2k}}& = -8 \frac{\sqrt3}{\pi}, \end{align}$

(1.31)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(208k^2+66k+23) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)48^{2k}}& = \frac{128}{\sqrt3\,\pi}, \end{align}$

(1.32)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(6722k^2-411k-152) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)(-63^2)^k}& = -195 \frac{\sqrt7}{\pi}, \end{align}$

(1.33)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(281591k^2-757041k-231992) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)(-63^2)^k}& = -274625 \frac{\sqrt7}{\pi}, \end{align}$

(1.34)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(560k^2-42k-11) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)12^{4k}}& = -24 \frac{\sqrt2}{\pi}, \end{align}$

(1.35)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(112k^2+114k+23) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)12^{4k}}& = \frac{256\sqrt2}{5\pi}, \end{align}$

(1.36)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(248k^2-18k-5) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)(-3\times2^{12})^k}& = - \frac{28}{\sqrt3\,\pi}, \end{align}$

(1.37)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(680k^2+1482k+337) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)(-3\times2^{12})^k}& = \frac{5488\sqrt3}{9\pi}, \end{align}$

(1.38)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(1144k^2-102k-19) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)(-2^{10}3^4)^k}& = - \frac{60}{\pi}, \end{align}$

(1.39)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(3224k^2+4026k+637) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)(-2^{10}3^4)^k}& = \frac{2000}{\pi}, \end{align}$

(1.40)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(7408k^2-754k-103) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)28^{4k}}& = - \frac{560\sqrt3}{3\pi}, \end{align}$

(1.41)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(3641424k^2+4114526k+493937) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)28^{4k}}& = 896000 \frac{\sqrt3}{\pi}, \end{align}$

(1.42)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(4744k^2-534k-55) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)(-2^{14}3^45)^k}& = - \frac{1932\sqrt5}{25\pi}, \end{align}$

(1.43)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(18446264k^2+20356230k+1901071) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)(-2^{14}3^45)^k}& = 66772496 \frac{\sqrt5}{25\pi}, \end{align}$

(1.44)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(413512k^2-50826k-3877) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)(-2^{10}21^4)^k}& = - \frac{12180}{\pi}, \end{align}$

(1.45)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(1424799848k^2+1533506502k+108685699) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)(-2^{10}21^4)^k}& = \frac{341446000}{\pi}, \end{align}$

(1.46)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(71312k^2-7746k-887) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)1584^{2k}}& = -840 \frac{\sqrt{11}}{\pi}, \end{align}$

(1.47)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(50678512k^2+56405238k+5793581) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)1584^{2k}}& = 5488000 \frac{\sqrt{11}}{\pi}, \end{align}$

(1.48)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(7329808k^2-969294k-54073) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)396^{4k}}& = -120120 \frac{\sqrt2}{\pi}, \end{align}$

(1.49)

$\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^\infty \frac{(2140459883152k^2+2259867244398k+119407598201) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)396^{4k}} \\&\qquad\qquad\qquad = 44\times1820^3 \frac{\sqrt2}{\pi}, \end{aligned} \end{equation}$

(1.50)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(164k^2-k-3) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)20^{3k}}& = - \frac{7\sqrt5}{2\pi}, \end{align}$

(1.51)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(2696k^2+206k+93) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)20^{3k}} & = \frac{686}{\sqrt5\,\pi}, \end{align}$

(1.52)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(220k^2-8k-3) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(-2^{15})^k} & = - \frac{7\sqrt2}{\pi}, \end{align}$

(1.53)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(836k^2-1048k-309) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(-2^{15})^k} & = - \frac{686\sqrt2}{\pi}, \end{align}$

(1.54)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(504k^2-11k-8) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(-15)^{3k}} & = - \frac{9\sqrt{15}}{\pi}, \end{align}$

(1.55)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(189k^2-11k-8) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(-15)^{3k}}& = - \frac{243\sqrt{15}}{35\pi}, \end{align}$

(1.56)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(516k^2-19k-7) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(2\times30^3)^k} & = - \frac{11\sqrt{15}}{2\pi}, \end{align}$

(1.57)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(3237k^2+1922k+491) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(2\times30^3)^k} & = \frac{3993\sqrt{15}}{10\pi}, \end{align}$

(1.58)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(684k^2-40k-7) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(-96)^{3k}}& = - \frac{9\sqrt6}{\pi}, \end{align}$

(1.59)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(2052k^2+2536k+379) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(-96)^{3k}}& = \frac{486\sqrt6}{\pi}, \end{align}$

(1.60)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(2556k^2-131k-29) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)66^{3k}}& = - \frac{63\sqrt{33}}{4\pi}, \end{align}$

(1.61)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(203985k^2+212248k+38083) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)66^{3k}}& = \frac{83349\sqrt{33}}{4\pi}, \end{align}$

(1.62)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(5812k^2-408k-49) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(-3\times160^3)^k}& = - \frac{253\sqrt{30}}{9\pi}, \end{align}$

(1.63)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(3471628k^2+3900088k+418289) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(-3\times160^3)^k}& = \frac{32388554\sqrt{30}}{135\pi}, \end{align}$

(1.64)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(35604k^2-2936k-233) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(-960)^{3k}}& = -189 \frac{\sqrt{15}}{\pi}, \end{align}$

(1.65)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(13983084k^2+15093304k+1109737) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(-960)^{3k}}& = \frac{4500846\sqrt{15}}{5\pi}, \end{align}$

(1.66)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(157752k^2-11243k-1304) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)255^{3k}}& = - \frac{513\sqrt{255}}{2\pi}, \end{align}$

(1.67)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(28240947k^2+31448587k+3267736) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)255^{3k}}& = \frac{45001899\sqrt{255}}{70\pi}, \end{align}$

(1.68)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(2187684k^2-200056k-11293) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(-5280)^{3k}} & = -1953 \frac{\sqrt{330}}{\pi}, \end{align}$

(1.69)

$\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^\infty \frac{(101740699836k^2+107483900696k+5743181813) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(-5280)^{3k}} \\&\qquad\qquad\qquad = \frac{4966100118\sqrt{330}}{5\pi}, \end{aligned} \end{equation}$

(1.70)

$\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^\infty \frac{(16444841148k^2-1709536232k-53241371) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(-640320)^{3k}} \\&\qquad\qquad\qquad = -1672209 \frac{\sqrt{10005}}{\pi}, \end{aligned} \end{equation}$

(1.71)

and

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

(1.72)

where

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

Recall that the Catalan numbers are given by

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

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

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

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

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

For example, (1.2) has the equivalent form

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

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

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

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

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

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

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

and

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

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

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

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

and

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

Now we state our second theorem.

Theorem 1.2. We have the identities

$\begin{align} \sum\limits_{k = 1}^\infty \frac{28k^2+31k+8}{(2k+1)^2k^3 \binom{2k}k^3}& = \frac{\pi^2-8}2, \end{align}$

(1.73)

$\begin{align} \sum\limits_{k = 1}^\infty \frac{42k^2+39k+8}{(2k+1)^3k^3 \binom{2k}k^3}& = \frac{9\pi^2-88}2, \end{align}$

(1.74)

$\begin{align} \sum\limits_{k = 1}^\infty \frac{(8k^2+5k+1)(-8)^k}{(2k+1)^2k^3 \binom{2k}k^3}& = 4-6G, \end{align}$

(1.75)

$\begin{align} \sum\limits_{k = 1}^\infty \frac{(30k^2+33k+7)(-8)^k}{(2k+1)^3k^3 \binom{2k}k^3}& = 54G-52, \end{align}$

(1.76)

$\begin{align} \sum\limits_{k = 1}^\infty \frac{(3k+1)16^k}{(2k+1)^2k^3 \binom{2k}k^3}& = \frac{\pi^2-8}2, \end{align}$

(1.77)

$\begin{align} \sum\limits_{k = 1}^\infty \frac{(4k+1)(-64)^k}{(2k+1)^2k^2 \binom{2k}k^3}& = 4-8G, \end{align}$

(1.78)

$\begin{align} \sum\limits_{k = 1}^\infty \frac{(4k+1)(-64)^k}{(2k+1)^3k^3 \binom{2k}k^3}& = 16G-16, \end{align}$

(1.79)

$\begin{align} \sum\limits_{k = 1}^\infty \frac{(2k^2-11k-3)8^k}{(2k+1)(3k+1)k^3 \binom{2k}k^2 \binom{3k}k}& = \frac{48-5\pi^2}2, \end{align}$

(1.80)

$\begin{align} \sum\limits_{k = 2}^\infty \frac{(178k^2-103k-39)8^k}{(k-1)(2k+1)(3k+1)k^3 \binom{2k}k^2 \binom{3k}k}& = \frac{1125\pi^2-11096}{36}, \end{align}$

(1.81)

$\begin{align} \sum\limits_{k = 1}^\infty \frac{(5k+1)(-27)^k}{(2k+1)(3k+1)k^2 \binom{2k}k^2 \binom{3k}k}& = 6-9K, \end{align}$

(1.82)

$\begin{align} \sum\limits_{k = 2}^\infty \frac{(45k^2+5k-2)(-27)^{k-1}}{(k-1)(2k+1)(3k+1)k^3 \binom{2k}k^2 \binom{3k}k}& = \frac{37-48K}{16}, \end{align}$

(1.83)

$\begin{align} \sum\limits_{k = 1}^\infty \frac{(98k^2-21k-8)81^k}{(2k+1)(4k+1)k^3 \binom{2k}k^2 \binom{4k}{2k}}& = 216-20\pi^2, \end{align}$

(1.84)

$\begin{align} \sum\limits_{k = 2}^\infty \frac{(1967k^2-183k-104)81^k}{(k-1)(2k+1)(4k+1)k^3 \binom{2k}k^2 \binom{4k}{2k}}& = \frac{20000\pi^2-190269}{120}, \end{align}$

(1.85)

$\begin{align} \sum\limits_{k = 1}^\infty \frac{(46k^2+3k-1)(-144)^k}{(2k+1)(4k+1)k^3 \binom{2k}k^2 \binom{4k}{2k}}& = 72- \frac{225}2K, \end{align}$

(1.86)

$\begin{align} \sum\limits_{k = 2}^\infty \frac{(343k^2+18k-16)(-144)^k}{(k-1)(2k+1)(4k+1)k^3 \binom{2k}k^2 \binom{4k}{2k}}& = \frac{9375K-7048}{10}, \end{align}$

(1.87)

where

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

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

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

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

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

For example, (1.77) has the following equivalent form

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

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

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

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

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

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

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

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

Now we state our third theorem.

Theorem 1.3. We have

$\begin{align} \sum\limits_{k = 0}^\infty \frac{7k+3}{24^k}S_k(1,-6)& = \frac{15}{\sqrt2\,\pi}, \end{align}$

(1.88)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{12k+5}{(-28)^k}S_k(1,7)& = \frac{6\sqrt7}{\pi}, \end{align}$

(1.89)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{84k+29}{80^k}S_k(1,-20)& = \frac{24\sqrt{15}}{\pi}, \end{align}$

(1.90)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{3k+1}{(-100)^k}S_k(1,25)& = \frac{25}{8\pi}, \end{align}$

(1.91)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{228k+67}{224^k}S_k(1,-56)& = \frac{80\sqrt7}{\pi}, \end{align}$

(1.92)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{399k+101}{(-676)^k}S_k(1,169)& = \frac{2535}{8\pi}, \end{align}$

(1.93)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{2604k+563}{2600^k}S_k(1,-650)& = \frac{850\sqrt{39}}{3\pi}, \end{align}$

(1.94)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{39468k+7817}{(-6076)^k}S_k(1,1519)& = \frac{4410\sqrt{31}}{\pi}, \end{align}$

(1.95)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{41667k+7879}{9800^k}S_k(1,-2450)& = \frac{40425\sqrt6}{4\pi}, \end{align}$

(1.96)

$\begin{align} \sum\limits_{k = 0}^\infty \frac{74613k+10711}{(-530^2)^k}S_k(1,265^2)& = \frac{1615175}{48\pi}. \end{align}$

(1.97)

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

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

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

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

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

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

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

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

(1.98)

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

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

(1.99)

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

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

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

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

(1.100)

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

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

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

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

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

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

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

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

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

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

(1.101)

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

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

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

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

(1.102)

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

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

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

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

(1.103)

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

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

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

2. Proofs of Theorems 1.1-1.3

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

$\begin{align} \sum\limits_{k = 0}^n \frac{((64-m)k^3-32k^2-16k+8) \binom{2k}k^3}{(2k-1)^2m^k} = & \frac{8(2n+1)}{m^n} \binom{2n}n^3, \end{align}$

(2.1)

$\begin{align} \sum\limits_{k = 0}^n \frac{((64-m)k^3-96k^2+48k-8) \binom{2k}k^3}{(2k-1)^3m^k} = & \frac 8{m^n} \binom{2n}n^3, \end{align}$

(2.2)

$\begin{align} \sum\limits_{k = 0}^n \frac{((108-m)k^3-54k^2-12k+6) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)m^k} = & \frac{6(3n+1)}{m^n} \binom{2n}n^2 \binom{3n}n, \end{align}$

(2.3)

$\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^n \frac{((108-m)k^3-(54+m)k^2-12k+6) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)m^k} \\&\qquad\qquad\qquad = \frac{6(3n+1)}{(n+1)m^n} \binom{2n}n^2 \binom{3n}n, \end{aligned} \end{equation}$

(2.4)

$\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^n \frac{((256-m)k^3-128k^2-16k+8) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)m^k} \\&\qquad\qquad\qquad = \frac{8(4n+1)}{m^n} \binom{2n}n^2 \binom{4n}{2n}, \end{aligned} \end{equation}$

(2.5)

$\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^n \frac{((256-m)k^3-(128+m)k^2-16k+8) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)m^k} \\&\qquad\qquad\qquad = \frac{8(4n+1)}{(n+1)m^n} \binom{2n}n^2 \binom{4n}{2n}, \end{aligned} \end{equation}$

(2.6)

$\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^n \frac{((1728-m)k^3-864k^2-48k+24) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)m^k} \\&\qquad\qquad\qquad = \frac{24(6n+1)}{m^n} \binom{2n}n \binom{3n}n \binom{6n}{3n}, \end{aligned} \end{equation}$

(2.7)

$\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^n \frac{((1728-m)k^3-(864+m)k^2-48k+24) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)m^k} \\&\qquad\qquad\qquad = \frac{24(6n+1)}{(n+1)m^n} \binom{2n}n \binom{3n}n \binom{6n}{3n}. \end{aligned} \end{equation}$

(2.8)

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

$\begin{gather} \binom{2n}n\sim \frac{4^n}{\sqrt{n\pi}},\ \ \binom{2n}n \binom{3n}n\sim \frac{\sqrt3\, 27^n}{2n\pi}, \end{gather}$

(2.9)

$\begin{gather} \binom{2n}n \binom{4n}{2n}\sim \frac{64^n}{\sqrt2\, n\pi},\ \ \binom{3n}n \binom{6n}n\sim \frac{432^n}{2n\pi}. \end{gather}$

(2.10)

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

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

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

Note that

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

and recall Bauer's series

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

So, we get

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

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

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

and hence

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

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

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

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

and hence

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

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

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

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

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

and hence

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

This proves $(1.71)$ .

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

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

and hence

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

Note that

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

Therefore, with the help of $(1.71)$ we get

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

This proves $(1.72)$ .

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

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

$\begin{align} \sum\limits_{k = 1}^n \frac{m^k((m-64)k^3-32k^2+16k+8)}{(2k+1)^2k^3 \binom{2k}k^3} = & \frac{m^{n+1}}{(2n+1)^2 \binom{2n}n^3}-m, \end{align}$

(2.11)

$\begin{align} \sum\limits_{k = 1}^n \frac{m^k((m-64)k^3-96k^2-48k-8)}{(2k+1)^3k^3 \binom{2k}k^3} = & \frac{m^{n+1}}{(2n+1)^3 \binom{2n}n^3}-m, \end{align}$

(2.12)

$\begin{align} \sum\limits_{k = 1}^n \frac{m^k((m-108)k^3-54k^2+12k+6)}{(2k+1)(3k+1)k^3 \binom{2k}k^2 \binom{3k}k} = & \frac{m^{n+1}}{(2n+1)(3n+1) \binom{2n}n^2 \binom{3n}n}-m, \end{align}$

(2.13)

$\begin{equation} \begin{aligned} &\sum\limits_{1 < k \leq n} \frac{m^k((m-108)k^3-(54+m)k^2+12k+6)}{(k-1)(2k+1)(3k+1)k^3 \binom{2k}k^2 \binom{3k}k} \\&\qquad\qquad = \frac{m^{n+1}}{n(2n+1)(3n+1) \binom{2n}n^2 \binom{3n}n}- \frac{m^2}{144}, \end{aligned} \end{equation}$

(2.14)

$\begin{equation} \begin{aligned} &\sum\limits_{k = 1}^n \frac{m^k((m-256)k^3-128k^2+16k+8)}{(2k+1)(4k+1)k^3 \binom{2k}k^2 \binom{4k}{2k}} \\&\qquad\qquad = \frac{m^{n+1}}{(2n+1)(4n+1) \binom{2n}n^2 \binom{4n}{2n}}-m, \end{aligned} \end{equation}$

(2.15)

$\begin{equation} \begin{aligned} &\sum\limits_{1 < k \leq n} \frac{m^k((m-256)k^3-(128+m)k^2+16k+8)}{(k-1)(2k+1)(4k+1)k^3 \binom{2k}k^2 \binom{4k}{2k}} \\&\qquad\qquad\quad = \frac{m^{n+1}}{n(2n+1)(4n+1) \binom{2n}n^2 \binom{4n}{2n}}- \frac{m^2}{360}. \end{aligned} \end{equation}$

(2.16)

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

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

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

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

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

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

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

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

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

Notice that

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

So we have

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

and hence (1.77) holds.

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

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

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

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

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

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

we see that

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

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

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

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

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

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

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

This proves (1.79).

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

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

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

and

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

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

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

and hence

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

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

with the aid of (1.82) we get

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

and hence $(1.83)$ follows.

Other identities in Theorem 1.2 can be proved similarly.

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

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

(2.17)

For $n\in{\Bbb N}$ we set

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

(2.18)

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

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

(2.19)

and

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

(2.20)

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

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

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

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

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

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

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

where

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

and

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

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

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

and

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

respectively. Therefore

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

and hence

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

satisfies the recurrence relation

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

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

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

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

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

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

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

(2.21)

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

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

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

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

with the aid of the Chu-Vandermonde identity. Therefore

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

This proves (2.21).

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

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

(2.22)

Proof. Let $n = k+l$ . Then

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

and

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

Hence

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

This proves (2.22).

To prove Theorem 1.3, we need an auxiliary theorem.

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

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

(2.23)

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

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

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

and similarly

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

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

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

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

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

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

and

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

$\begin{align*} \leq&\sum\limits_{N/2 < l \leq N}2l^2(2l+1) \left( \frac{16}m \right)^lP_l(3) \\ \leq&2\sum\limits_{l > N/2}l^2(2l+1)P_l(3) \left( \frac{16}m \right)^l. \end{align*}$

Recall that

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

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

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

converges. Thus

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

and hence by the above we have

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

and

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

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

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

(2.24)

and

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

(2.25)

In view of (2.25) and (2.20),

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

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

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

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

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

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

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

Therefore

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

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

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

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

3. New series involving $T_n(b,c)$ for $1/\pi$ related to types Ⅰ-Ⅷ

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 $8$ .

4. Congruences related to Theorem 1.3

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 $p$ .

5. Series for $1/\pi$ involving $T_n(b,c)$ and $Z_n=\sum_{k=0}^n \binom nk \binom{2k}k \binom{2(n-k)}{n-k}$

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.

6. Series for $1/\pi$ involving $T_k(b,c)$ and the Franel numbers

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.

7. Series for $1/\pi$ involving $T_n(b,c)$ and $g_n=\sum_{k=0}^n \binom nk^2 \binom{2k}k$

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

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

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

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).

8. Series and congruences involving $T_n(b,c)$ and $\beta_n=\sum_{k=0}^n \binom nk^2 \binom{n+k}k$

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)

9. Series and congruences involving $w_n=\sum_{k=0}^{\lfloor n/3\rfloor}(-1)^k3^{n-3k} \binom n{3k} \binom{3k}k \binom{2k}k$ and $T_n(b,c)$

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.

10. Series for $\pi$ involving $T_n$ and related congruences

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

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

Acknowledgment

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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Learning deep forest for face anti-spoofing: An alternative to the neural network against adversarial attacks

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Abstract

1. Introduction and our main results

2. Proofs of Theorems 1.1-1.3

3. New series involving $T_n(b,c)$ for $1/\pi$ related to types Ⅰ-Ⅷ

4. Congruences related to Theorem 1.3

5. Series for $1/\pi$ involving $T_n(b,c)$ and $Z_n=\sum_{k=0}^n \binom nk \binom{2k}k \binom{2(n-k)}{n-k}$

6. Series for $1/\pi$ involving $T_k(b,c)$ and the Franel numbers

7. Series for $1/\pi$ involving $T_n(b,c)$ and $g_n=\sum_{k=0}^n \binom nk^2 \binom{2k}k$

8. Series and congruences involving $T_n(b,c)$ and $\beta_n=\sum_{k=0}^n \binom nk^2 \binom{n+k}k$

9. Series and congruences involving $w_n=\sum_{k=0}^{\lfloor n/3\rfloor}(-1)^k3^{n-3k} \binom n{3k} \binom{3k}k \binom{2k}k$ and $T_n(b,c)$

10. Series for $\pi$ involving $T_n$ and related congruences

Acknowledgment

References

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Electronic Research Archive

Learning deep forest for face anti-spoofing: An alternative to the neural network against adversarial attacks

Related Papers:

Abstract

1. Introduction and our main results

2. Proofs of Theorems 1.1-1.3

3. New series involving T_n(b,c) for 1/\pi related to types Ⅰ-Ⅷ

4. Congruences related to Theorem 1.3

5. Series for 1/\pi involving T_n(b,c) and Z_n=\sum_{k=0}^n \binom nk \binom{2k}k \binom{2(n-k)}{n-k}

6. Series for 1/\pi involving T_k(b,c) and the Franel numbers

7. Series for 1/\pi involving T_n(b,c) and g_n=\sum_{k=0}^n \binom nk^2 \binom{2k}k

8. Series and congruences involving T_n(b,c) and \beta_n=\sum_{k=0}^n \binom nk^2 \binom{n+k}k

9. Series and congruences involving w_n=\sum_{k=0}^{\lfloor n/3\rfloor}(-1)^k3^{n-3k} \binom n{3k} \binom{3k}k \binom{2k}k and T_n(b,c)

10. Series for \pi involving T_n and related congruences

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3. New series involving $T_n(b,c)$ for $1/\pi$ related to types Ⅰ-Ⅷ

5. Series for $1/\pi$ involving $T_n(b,c)$ and $Z_n=\sum_{k=0}^n \binom nk \binom{2k}k \binom{2(n-k)}{n-k}$

6. Series for $1/\pi$ involving $T_k(b,c)$ and the Franel numbers

7. Series for $1/\pi$ involving $T_n(b,c)$ and $g_n=\sum_{k=0}^n \binom nk^2 \binom{2k}k$

8. Series and congruences involving $T_n(b,c)$ and $\beta_n=\sum_{k=0}^n \binom nk^2 \binom{n+k}k$

9. Series and congruences involving $w_n=\sum_{k=0}^{\lfloor n/3\rfloor}(-1)^k3^{n-3k} \binom n{3k} \binom{3k}k \binom{2k}k$ and $T_n(b,c)$

10. Series for $\pi$ involving $T_n$ and related congruences