Proof of Sun's conjectural supercongruence involving Catalan numbers

Ji-Cai Liu; Ji-Cai Liu

doi:10.3934/era.2020054

Electronic Research Archive

2020, Volume 28, Issue 2: 1023-1030. doi: 10.3934/era.2020054

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Proof of Sun's conjectural supercongruence involving Catalan numbers

Ji-Cai Liu

Department of Mathematics, Wenzhou University, Wenzhou 325035, China

Received: 01 January 2020 Revised: 01 April 2020
Primary: 11A07, 11B65; Secondary: 05A19, 33C20

We confirm a conjectural supercongruence involving Catalan numbers, which is one of the 100 selected open conjectures on congruences of Sun. The proof makes use of hypergeometric series identities and symbolic summation method.

Keywords:

Citation: Ji-Cai Liu. Proof of Sun's conjectural supercongruence involving Catalan numbers[J]. Electronic Research Archive, 2020, 28(2): 1023-1030. doi: 10.3934/era.2020054

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Abstract

1. Introduction

In 2003, Rodriguez-Villegas [14] conjectured the following four supercongruences associated to certain elliptic curves:

$\begin{align*} \sum\limits_{k = 0}^{p-1}\frac{{2k\choose k}^2}{16^k}\equiv \left(\frac{-1}{p}\right) \pmod{p^2}, \quad \sum\limits_{k = 0}^{p-1}\frac{{2k\choose k}{3k\choose k}}{27^k}\equiv \left(\frac{-3}{p}\right) \pmod{p^2}, \\ \sum\limits_{k = 0}^{p-1}\frac{{2k\choose k}{4k\choose 2k}}{64^k}\equiv \left(\frac{-2}{p}\right) \pmod{p^2}, \quad \sum\limits_{k = 0}^{p-1}\frac{{3k\choose k}{6k\choose 3k}}{432^k}\equiv \left(\frac{-1}{p}\right) \pmod{p^2}, \end{align*}$

where $p\ge 5$ is a prime and $\left(\frac{\cdot}{p}\right)$ denotes the Legendre symbol. These four supercongruences were first proved by Mortenson [12,13] by using the Gross-Koblitz formula. Guo, Pan and Zhang [3] established some interesting $q$ -analogues of the above four supercongruences. For more $q$ -analogues of congruences, one can refer to [1,2,4,5,10].

Recall that the Euler numbers are defined as

$\begin{align*} \frac{2}{e^x+e^{-x}} = \sum\limits_{n = 0}^{\infty}E_n\frac{x^n}{n!}, \end{align*}$

and the $n$ th Catalan number is given by

$\begin{align*} C_n = \frac{1}{n+1}{2n\choose n}, \end{align*}$

which plays an important role in various counting problems. We refer to [17] for many different combinatorial interpretations of the Catalan numbers.

In 2016, Z.-H. Sun [18] proved that for any prime $p\ge 5$ ,

$\begin{align*} \sum\limits_{k = 0}^{p-1}\frac{{2k\choose k}C_{2k}}{64^k} \equiv (-1)^{\frac{p-1}{2}}-3p^2E_{p-3}\pmod{p^3}, \end{align*}$

which was originally conjectured by Z.-W. Sun [19].

Mao and Z.-W. Sun [11] showed that for any prime $p\ge 5$ ,

$\begin{align} \sum\limits_{k = 0}^{(p-1)/2}\frac{{2k\choose k}C_{2k}}{64^k}\equiv (-1)^{\frac{p-1}{2}}2^{p-1}\pmod{p^2}. \end{align}$

(1)

Z.-W. Sun [22,Conjecture 11] also conjectured an extension of (1) as follows.

Conjecture 1.1 (Sun, 2019). For any prime $p\ge 5$ , we have

$\begin{align} \sum\limits_{k = 0}^{(p-1)/2}\frac{{2k\choose k}C_{2k}}{64^k} \equiv (-1)^{\frac{p-1}{2}}\left(2^{p-1}-(2^{p-1}-1)^2\right)\pmod{p^3}. \end{align}$

(2)

The main purpose of the paper is to prove (2). Our proof is based on hypergeometric series identities and symbolic summation method.

Theorem 1.2. The supercongruence (2) is true.

We establish two preliminary results in the next section. The proof of Theorem 1.2 will be given in Section 3.

2. Preliminary results

In order to prove Theorem 1.2, we need the following two key results.

Proposition 2.1. For any prime $p\ge 5$ , we have

$\begin{align} &\sum\limits_{k = 0}^{(p-1)/2}\frac{{2k\choose k}}{(2k-1)^2 4^k}\equiv (-1)^{\frac{p-1}{2}}\left(2^{p-1}-q_p(2)\right)\pmod{p^2}, \end{align}$

(3)

$\begin{align} &\sum\limits_{k = 0}^{(p-1)/2}\frac{{2k\choose k}^2}{(2k-1)^3 16^k}\equiv 2-2q_p(2)-p\left(q_p(2)^2-4q_p(2)+3\right)\pmod{p^2}, \end{align}$

(4)

where $q_p(2)$ is the Fermat quotient $(2^{p-1}-1)/p$ .

Remark. Z.-W. Sun [20,(1.7)] and [21,(1.7),(3.3),(3.4)] has proved the following closely related results:

$\begin{align*} &\sum\limits_{k = 0}^{(p-3)/2}\frac{{2k\choose k}}{(2k+1)^24^k}\equiv (-1)^{\frac{p+1}{2}} \frac{q_p(2)^2}{2}\pmod{p}, \\ &\sum\limits_{k = 0}^{(p-3)/2}\frac{{2k\choose k}^2}{(2k+1)16^k} \equiv -2q_p(2)-pq_p(2)^2+\frac{5p^2}{12}B_{p-3}\pmod{p^3}, \\ &\sum\limits_{k = 0}^{(p-3)/2}\frac{{2k\choose k}^2}{(2k+1)^216^k}\equiv -2q_p(2)^2+\frac{2p}{3}q_p(2)^3 -\frac{p}{6}B_{p-3}\pmod{p^2}, \\ &\sum\limits_{k = 0}^{(p-3)/2}\frac{{2k\choose k}^2}{(2k+1)^3 16^k}\equiv -\frac{4}{3}q_p(2)^3-\frac{1}{6}B_{p-3}\pmod{p}, \end{align*}$

where the Bernoulli numbers are given by

$\begin{align*} \frac{x}{e^x-1} = \sum\limits_{n = 0}^{\infty}B_n\frac{x^n}{n!}. \end{align*}$

Before proving Proposition 2.1, we establish the following lemma.

Lemma 2.2. For any integer $n\ge 2$ , we have

$\begin{align} \sum\limits_{k = 0}^n\frac{(-n)_k (n-1)_k}{(1)_k\left(\frac{1}{2}\right)_k} = \frac{(-1)^{n-1}}{2n-1}, \end{align}$

(5)

$\begin{align} \sum\limits_{k = 0}^n\frac{(-n)_k (n-1)_k \left(-\frac{1}{2}\right)_k}{(1)_k^2\left(\frac{1}{2}\right)_k} = \frac{4n(n-1)}{2n-1}, \end{align}$

(6)

where $(a)_0 = 1$ and $(a)_k = a(a+1)\cdots (a+k-1)$ for $k\ge 1$ .

Proof. Recall Gauss' theorem [16,(1.7.6),page 28]:

$\begin{align} _2{F_1} \left[ {\begin{array}{*{20}{c}} a&b\\ {}&c \end{array};1} \right] = \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}, \end{align}$

(7)

provided that $\Re(c-a-b)>0$ . Letting $a = -n, b = n-1$ and $c = \frac{1}{2}$ in (7) gives

$\begin{align*} _2{F_1} \left[ {\begin{array}{*{20}{c}} { - n}&{n - 1}\\ {}&{\frac{1}{2}} \end{array};1} \right] = \frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{3}{2}\right)} {\Gamma\left(\frac{1}{2}+n\right)\Gamma\left(\frac{3}{2}-n\right)} = \frac{(-1)^{n-1}}{2n-1}, \end{align*}$

which is (5).

Also, we have the following transformation formula of hypergeometric series [16,(2.5.11),page 76]:

$\begin{align} {_3{F_2} }\left[ {\begin{array}{*{20}{c}} a&b&{ - n}\\ {}&e&f \end{array};1} \right] = \frac{(e-a)_n (f-a)_n}{(e)_n(f)_n}\\ \times {_3{F_2} }\left[ {\begin{array}{*{20}{c}} {1 - s}&a&{ - n}\\ {}&{1 + a - e - n}&{1 + a - f - n} \end{array};1} \right] , \end{align}$

(8)

where $s = e+f-a-b+n$ . Letting $a = n-1, b = -\frac{1}{2}, e = x$ and $f = \frac{3}{2}-x$ in (8) yields

$\begin{align} {_3{F_2} }\left[ {\begin{array}{*{20}{c}} {n - 1}&{ - \frac{1}{2} }&{- n}\\ {}&x&{\frac{3}{2} - x} \end{array};1} \right] = \frac{(x+1-n)_n \left(\frac{5}{2}-x-n\right)_n}{(x)_n\left(\frac{3}{2}-x\right)_n}\\ \times {_3{F_2} }\left[ {\begin{array}{*{20}{c}} { - 2}&{n - 1}&{ - n}\\ {}&{ - x}&{x - \frac{3}{2}} \end{array}} ;1\right]. \end{align}$

(9)

Furthermore, we can evaluate the terminating hypergeometric series on the right-hand side of (9):

${_3{F_2} }\left[ {\begin{array}{*{20}{c}} { - 2}&{n - 1}&{ - n}\\ {}&{ - x}&{x - \frac{3}{2}} \end{array};1} \right] \\ = \frac{4x^4-12x^3+(-8n^2+8n+11)x^2+(12n^2-12n-3)x+4n(n-1)(n^2-n-1)}{x(x-1)(2x-1)(2x-3)}.$

It follows that

$\begin{align} {_3{F_2} }\left[ {\begin{array}{*{20}{c}} {n - 1}&{ - \frac{1}{2}}&{ - n}\\ {}&x&{\frac{3}{2} - x} \end{array};1} \right] = \frac{(x+1-n)_n \left(\frac{5}{2}-x-n\right)_n}{(x)_n\left(\frac{3}{2}-x\right)_n}\\ \times\frac{4x^4-12x^3+(-8n^2+8n+11)x^2+(12n^2-12n-3)x+4n(n-1)(n^2-n-1)}{x(x-1)(2x-1)(2x-3)}. \end{align}$

(10)

Letting $x\to 1$ on both sides of (10) and noting that

$\begin{align*} &\lim\limits_{x\to 1}\frac{4x^4-12x^3+(-8n^2+8n+11)x^2+(12n^2-12n-3)x+4n(n-1)(n^2-n-1)}{x(2x-1)(2x-3)}\\ & = -4n^2(n-1)^2, \end{align*}$

and

$\begin{align*} \lim\limits_{x\to 1}\frac{(x+1-n)_n \left(\frac{5}{2}-x-n\right)_n}{(x-1)(x)_n\left(\frac{3}{2}-x\right)_n} = -\frac{1}{n(n-1)(2n-1)}, \end{align*}$

we arrive at

$\begin{align*} {_3{F_2} }\left[ {\begin{array}{*{20}{c}} {n - 1}&{ - \frac{1}{2}}&{ - n}\\ {}&1&{\frac{1}{2}} \end{array};1} \right] = \frac{4n(n-1)}{2n-1}, \end{align*}$

which proves (6).

Proof of (3). We can rewrite (5) as

$\begin{align} \sum\limits_{k = 0}^{n-1}\frac{(-n)_k (n-1)_k}{(1)_k\left(\frac{1}{2}\right)_k}& = \frac{(-1)^{n-1}}{2n-1} -\frac{(-n)_n (n-1)_n}{(1)_n\left(\frac{1}{2}\right)_n}\\ & = \frac{(-1)^{n-1}}{2n-1}\left(1+4^{n-1}(2n-2)\right)\\ & = (-1)^{n-1}\left(2^{2n-2}-\frac{2^{2n-2}-1}{2n-1}\right). \end{align}$

(11)

Letting $n = \frac{p+1}{2}$ in (11) gives

$\begin{align*} \sum\limits_{k = 0}^{(p-1)/2}\frac{\left(\frac{-1-p}{2}\right)_k \left(\frac{-1+p}{2}\right)_k} {(1)_k \left(\frac{1}{2}\right)_k} = (-1)^{\frac{p-1}{2}}\left(2^{p-1}-q_p(2)\right). \end{align*}$

Since for $0\le k \le \frac{p-1}{2}$ ,

$\begin{align} \left(\frac{-1-p}{2}\right)_k \left(\frac{-1+p}{2}\right)_k\equiv \left(-\frac{1}{2}\right)_k^2\pmod{p^2}, \end{align}$

(12)

we have

$\begin{align} \sum\limits_{k = 0}^{(p-1)/2}\frac{ \left(-\frac{1}{2}\right)_k^2} {(1)_k \left(\frac{1}{2}\right)_k}\equiv (-1)^{\frac{p-1}{2}}\left(2^{p-1}-q_p(2)\right)\pmod{p^2}. \end{align}$

(13)

Note that

$\begin{align} &\frac{\left(\frac{1}{2}\right)_k}{(1)_k} = \frac{{2k\choose k}}{4^k}, \end{align}$

(14)

$\begin{align} &\frac{\left(-\frac{1}{2}\right)_k}{\left(\frac{1}{2}\right)_k} = \frac{1}{1-2k}. \end{align}$

(15)

Then the proof of (3) follows from (13)–(15).

Proof of (4). We can rewrite (6) as

$\begin{align} \sum\limits_{k = 0}^{n-1}\frac{(-n)_k (n-1)_k \left(-\frac{1}{2}\right)_k}{(1)_k^2\left(\frac{1}{2}\right)_k} & = \frac{4n(n-1)}{2n-1}-\frac{(-n)_n (n-1)_n \left(-\frac{1}{2}\right)_n}{(1)_n^2\left(\frac{1}{2}\right)_n}\\ & = \frac{1}{2n-1}\left(4n(n-1)+(-1)^n{2n-2\choose n}\right). \end{align}$

(16)

Letting $n = \frac{p+1}{2}$ in (16) and using (12), we obtain

$\begin{align*} \sum\limits_{k = 0}^{(p-1)/2}\frac{\left(-\frac{1}{2}\right)_k^3}{(1)_k^2\left(\frac{1}{2}\right)_k} \equiv \frac{1}{p}\left(p^2-1+(-1)^{\frac{p+1}{2}}{p-1\choose \frac{p+1}{2}}\right)\pmod{p^2}. \end{align*}$

For $0\le k\le p-1$ , we have

$\begin{align} {p-1\choose k}&\equiv (-1)^k\left(1-p\sum\limits_{i = 1}^k\frac{1}{i}+p^2\sum\limits_{1\le i < j\le k}\frac{1}{ij}\right)\\ & = (-1)^k\left(1-pH_k+\frac{p^2}{2}\left(H_k^2-H_k^{(2)}\right)\right)\pmod{p^3}, \end{align}$

(17)

where

$\begin{align*} H_k^{(r)} = \sum\limits_{j = 1}^k\frac{1}{j^r}, \end{align*}$

with the convention that $H_k = H_k^{(1)}$ . It follows that

$\begin{align} \sum\limits_{k = 0}^{(p-1)/2}\frac{\left(-\frac{1}{2}\right)_k^3}{(1)_k^2\left(\frac{1}{2}\right)_k} &\equiv \frac{p}{2}\left(H_{\frac{p+1}{2}}^2-H_{\frac{p+1}{2}}^{(2)}+2\right)-H_{\frac{p+1}{2}}\\ & = \frac{p}{2}\left(H_{\frac{p-1}{2}}^2+4H_{\frac{p-1}{2}}-H_{\frac{p-1}{2}}^{(2)}+6\right)-H_{\frac{p-1}{2}}-2 \pmod{p^2}. \end{align}$

(18)

By [7,(41)] and [19,Lemma 2.4], we have

$\begin{align} H_{\frac{p-1}{2}}\equiv -2q_p(2)+pq_p(2)^2\pmod{p^2}, \end{align}$

(19)

and

$\begin{align} H_{\frac{p-1}{2}}^{(2)}\equiv 0\pmod{p}. \end{align}$

(20)

Substituting (19) and (20) into (18) gives

$\begin{align} \sum\limits_{k = 0}^{(p-1)/2}\frac{\left(-\frac{1}{2}\right)_k^3}{(1)_k^2\left(\frac{1}{2}\right)_k} \equiv 2q_p(2)-2+p\left(q_p(2)^2-4q_p(2)+3\right)\pmod{p^2}. \end{align}$

(21)

Finally, applying (14) and (15) to the left-hand side of (21), we reach

$\begin{align*} \sum\limits_{k = 0}^{(p-1)/2}\frac{{2k\choose k}^2}{(2k-1)^3 16^k}\equiv 2-2q_p(2)-p\left(q_p(2)^2-4q_p(2)+3\right)\pmod{p^2}, \end{align*}$

as desired.

3. Proof of Theorem 1.2

Lemma 3.1. For any non-negative integer $n$ , we have

$\begin{align} &\sum\limits_{k = 0}^n\frac{\left(-n\right)_k\left(n+1\right)_k\left(\frac{1}{4}\right)_k \left(\frac{3}{4}\right)_k} {(1)_k^2\left(\frac{1}{2}\right)_k\left(\frac{3}{2}\right)_k} = \frac{{2n\choose n}}{4^n}, \end{align}$

(22)

and

$\begin{align} \sum\limits_{k = 0}^n\frac{\left(-n\right)_k\left(n+1\right)_k\left(\frac{1}{4}\right)_k \left(\frac{3}{4}\right)_k} {(1)_k^2\left(\frac{1}{2}\right)_k\left(\frac{3}{2}\right)_k}\sum\limits_{j = 1}^k\frac{1}{(2j-1)^2}\\ = -\frac{{2n\choose n}}{4^n}\left(3+\sum\limits_{k = 1}^n\frac{1}{(2k-1)^2}\right)+\frac{2}{2n+1}\sum\limits_{k = 0}^n \frac{{2k\choose k}}{(2k-1)^2 4^k}\\ -\frac{4^n}{(2n+1){2n\choose n}}\sum\limits_{k = 0}^n \frac{{2k\choose k}^2}{(2k-1)^3 16^k}. \end{align}$

(23)

Proof. Recall that (see [16,(2.4.2.2),page 65])

$\begin{align} _4{F_3} \left[ {\begin{array}{*{20}{c}} d&{1 + f - g}&{\frac{f}{2}}&{\frac{{f + 1}}{2}}\\ {}&{1 + f}&{\frac{{1 + f + d + g}}{2}}&{1 + \frac{{f + d - g}}{2}} \end{array};1} \right] = \frac{\Gamma(g-f)\Gamma(g-d)}{\Gamma(g)\Gamma(g-f-d)}. \end{align}$

(24)

Letting $d = -n, f = \frac{1}{2}$ and $g = -n+\frac{1}{2}$ in (24), we obtain

$\begin{align*} _4{F_3} \left[ {\begin{array}{*{20}{c}} { - n}&{n + 1}&{\frac{1}{4}}&{\frac{3}{4}}\\ {}&1&{\frac{1}{2}}&{\frac{3}{2}} \end{array};1} \right] = \frac{\Gamma(-n)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(-n+\frac{1}{2}\right)\Gamma(0)} = \frac{{2n\choose n}}{4^n}, \end{align*}$

which is (22).

On the other hand, (23) can be discovered and proved by symbolic summation package $\mathsf{ Sigma}$ due to Schneider [15]. One can refer to [9] for the same approach to finding and proving identities of this type.

Proof of (2). Recall that (see [8,(4.4)])

$\begin{align} \left(\frac{1+p}{2}\right)_k\left(\frac{1-p}{2}\right)_k \equiv \left(\frac{1}{2}\right)_k^2\left(1-p^2\sum\limits_{j = 1}^k\frac{1}{(2j-1)^2}\right) \pmod{p^4}. \end{align}$

(25)

Letting $n = \frac{p-1}{2}$ in (22) and using (25), we obtain

$\begin{align} \sum\limits_{k = 0}^{(p-1)/2}\frac{\left(\frac{1}{2}\right)_k\left(\frac{1}{4}\right)_k \left(\frac{3}{4}\right)_k}{(1)_k^2\left(\frac{3}{2}\right)_k} &\equiv \frac{1}{2^{p-1}}{p-1\choose \frac{p-1}{2}}\\ &+p^2\sum\limits_{k = 0}^{(p-1)/2}\frac{\left(\frac{1}{2}\right)_k\left(\frac{1}{4}\right)_k \left(\frac{3}{4}\right)_k}{(1)_k^2\left(\frac{3}{2}\right)_k}\sum\limits_{j = 1}^k\frac{1}{(2j-1)^2} \pmod{p^4}, \end{align}$

(26)

where we have utilized the fact $\left(\frac{1}{4}\right)_k \left(\frac{3}{4}\right)_k/\left(\frac{3}{2}\right)_k \in \mathbb{Z}_p$ for $0\le k\le \frac{p-1}{2}$ .

From (25), we deduce that

$\begin{align} \left(\frac{1+p}{2}\right)_k\left(\frac{1-p}{2}\right)_k \equiv \left(\frac{1}{2}\right)_k^2\pmod{p^2}. \end{align}$

(27)

Letting $n = \frac{p-1}{2}$ in (23) and using (27) gives

$\begin{align} &\sum\limits_{k = 0}^{(p-1)/2}\frac{\left(\frac{1}{2}\right)_k\left(\frac{1}{4}\right)_k \left(\frac{3}{4}\right)_k}{(1)_k^2\left(\frac{3}{2}\right)_k}\sum\limits_{j = 1}^k\frac{1}{(2j-1)^2}\\ &\equiv -\frac{1}{2^{p-1}}{p-1\choose \frac{p-1}{2}}\left(3+\sum\limits_{k = 1}^{(p-1)/2}\frac{1}{(2k-1)^2}\right)+\frac{2}{p}\sum\limits_{k = 0}^{(p-1)/2} \frac{{2k\choose k}}{(2k-1)^2 4^k}\\ &-\frac{2^{p-1}}{p{p-1\choose \frac{p-1}{2}}}\sum\limits_{k = 0}^{(p-1)/2} \frac{{2k\choose k}^2}{(2k-1)^3 16^k}\pmod{p^2}. \label{c-6} \end{align}$

(28)

Substituting (28) into (26) yields

$\begin{align} &\sum\limits_{k = 0}^{(p-1)/2}\frac{\left(\frac{1}{2}\right)_k\left(\frac{1}{4}\right)_k \left(\frac{3}{4}\right)_k}{(1)_k^2\left(\frac{3}{2}\right)_k} \equiv \frac{1}{2^{p-1}}{p-1\choose \frac{p-1}{2}} -\frac{p^2}{2^{p-1}}{p-1\choose \frac{p-1}{2}}\left(3+\sum\limits_{k = 1}^{(p-1)/2}\frac{1}{(2k-1)^2}\right)\\ &+2p\sum\limits_{k = 0}^{(p-1)/2}\frac{{2k\choose k}}{(2k-1)^2 4^k} -\frac{2^{p-1}p}{{p-1\choose \frac{p-1}{2}}}\sum\limits_{k = 0}^{(p-1)/2} \frac{{2k\choose k}^2}{(2k-1)^3 16^k}\pmod{p^4}. \end{align}$

(29)

Furthermore, by (17), (19) and (20) we have

$\begin{align} {p-1\choose \frac{p-1}{2}}&\equiv (-1)^{\frac{p-1}{2}}\left(1-pH_{\frac{p-1}{2}}+\frac{p^2}{2}\left(H_{\frac{p-1}{2}}^2-H_{\frac{p-1}{2}}^{(2)}\right)\right)\\ &\equiv (-1)^{\frac{p-1}{2}}\left(1+2pq_p(2)+p^2q_p(2)^2\right)\pmod{p^3}. \end{align}$

(30)

By (20) and the Wolstenholme's theorem [6,page 114], we have

$\begin{align} \sum\limits_{k = 1}^{(p-1)/2}\frac{1}{(2k-1)^2} = H_{p-1}^{(2)}-\frac{1}{4}H_{\frac{p-1}{2}}^{(2)} \equiv 0\pmod{p}. \end{align}$

(31)

Setting $2^{p-1} = a$ and $q_p(2) = (a-1)/p$ , and then substituting (3), (4), (30) and (31) into (29), we arrive at

$\begin{align*} &\sum\limits_{k = 0}^{(p-1)/2}\frac{\left(\frac{1}{2}\right)_k\left(\frac{1}{4}\right)_k \left(\frac{3}{4}\right)_k}{(1)_k^2\left(\frac{3}{2}\right)_k} \equiv (-1)^{\frac{p-1}{2}}\left(\frac{a^3-2a^2+4a-2}{2a-1}+\frac{3(a-1)^2p^2}{a(2a-1)}\right)\\ & = (-1)^{\frac{p-1}{2}}\left(a-(a-1)^2+\frac{3(a-1)^3}{2a-1}+\frac{3(a-1)^2p^2}{a(2a-1)}\right)\pmod{p^3}. \end{align*}$

By the Fermat's little theorem, we have $a-1\equiv 0\pmod{p}$ , and so

(32)

Note that

$\begin{align} &\frac{\left(\frac{1}{4}\right)_k\left(\frac{3}{4}\right)_k}{(1)_k^2} = \frac{{4k\choose 2k}{2k\choose k}}{64^k}, \end{align}$

(33)

$\begin{align} &\frac{\left(\frac{1}{2}\right)_k}{\left(\frac{3}{2}\right)_k} = \frac{1}{2k+1}. \end{align}$

(34)

Then the proof of (2) follows from (32)–(34).

Acknowledgments

The author would like to thank Professor Zhi-Wei Sun and the anonymous referees for helpful comments which made the paper more readable.

References

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Proof of Sun's conjectural supercongruence involving Catalan numbers

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1. Introduction

2. Preliminary results

3. Proof of Theorem 1.2

Acknowledgments

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Proof of Sun's conjectural supercongruence involving Catalan numbers

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Abstract

1. Introduction

2. Preliminary results

3. Proof of Theorem 1.2

Acknowledgments

References

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