NEW GENERALISATIONS OF VAN HAMME’S (G.2) SUPERCONGRUENCE

Abstract

Swisher [‘On the supercongruence conjectures of van Hamme’, Res. Math. Sci. 2 (2015), Article no. 18] and He [‘Supercongruences on truncated hypergeometric series’, Results Math. 72 (2017), 303–317] independently proved that Van Hamme’s (G.2) supercongruence holds modulo $p^4$ for any prime $p\equiv 1\pmod {4}$ . Swisher also obtained an extension of Van Hamme’s (G.2) supercongruence for $p\equiv 3 \pmod 4$ and $p>3$ . In this note, we give new one-parameter generalisations of Van Hamme’s (G.2) supercongruence modulo $p^3$ for any odd prime p. Our proof uses the method of ‘creative microscoping’ introduced by Guo and Zudilin [‘A q-microscope for supercongruences’, Adv. Math. 346 (2019), 329–358].

1 Introduction

In his first letter to Hardy in 1913, Ramanujan mentioned the following formula (see [1, page 25, (2)]):

(1.1) $$ \begin{align} \sum_{k=0}^\infty(8k+1)\frac{(\frac{1}{4})_k^4}{k!^4} =\frac{2\sqrt{2}}{\sqrt{\pi}\,\Gamma(\frac 34)^2}. \end{align} $$

Here $(a)_n=a(a+1)\cdots (a+n-1)$ is the rising factorial and $\Gamma (x) $ is the Gamma function. Ramanujan did not give a proof of (1.1) and the first proof was given by Hardy [7]. In 1997, Van Hamme [12] conjectured 13 p-adic analogues of Ramanujan’s series, including

(1.2) $$ \begin{align} &\sum_{k=0}^{(p-1)/4}(8k+1)\frac{(\frac{1}{4})_k^4}{k!^4} \equiv p\frac{\Gamma_p(\frac 12)\Gamma_p(\frac 14)}{\Gamma_p(\frac 34)} \pmod{p^3} \quad\text{for}\ p\equiv 1 \pmod 4 \end{align} $$

(marked (G.2) in Van Hamme’s list). Here and throughout the paper, p always denotes an odd prime and $\Gamma _p(x)$ stands for Morita’s p-adic Gamma function [10]. Swisher [11] and He [8] independently showed that (1.2) holds modulo the stronger power $p^4$ .

We shall give a generalisation of (1.2): for $p\equiv 1 \pmod 4$ and $0\leqslant s\leqslant (p-1)/4$ ,

(1.3) $$ \begin{align} \sum_{k=s}^{(p-1)/4} (8k+1)\frac{(\frac{1}{4})_{k-s}(\frac{1}{4})_{k+s}(\frac{1}{4})_k^2} {(k-s)!(k+s)!k!^2} \equiv (p+4s)\frac{(\frac{1}{4})_s^2(\frac{1}{4})_{(p-1)/4+s} (\frac{1}{2})_{(p-1)/4-s}} {s!^2(1)_{(p-1)/4+s}(\frac{1}{4})_{(p-1)/4-s}} \pmod{p^3}. \end{align} $$

When $s=0$ , the right-hand side of (1.3) reduces to $p(\tfrac 12)_{(p-1)/4}/(1)_{(p-1)/4}$ , which is congruent to the right-hand side of (1.2) modulo $p^3$ (see [9]). Thus, the supercongruence (1.3) is indeed a generalisation of (1.2). A similar extension of the (A.2) supercongruence of Van Hamme was recently given by Guo [3].

We shall prove (1.3) by establishing the following q-supercongruence.

Theorem 1.1. Let $n\equiv 1\pmod 4$ be an integer greater than $1$ and let $0\leqslant s\leqslant(n-1)/4$ . Then

(1.4) $$ \begin{align} &\sum_{k=s}^{(n-1)/4} [8k+1]\frac{(q;q^4)_{k-s}(q;q^4)_{k+s}(q;q^4)_k^2} {(q^4;q^4)_{k-s}(q^4;q^4)_{k+s}(q^4;q^4)_k^2} q^{2k} \notag\\[6pt] &\quad\equiv [n+4s] \frac{(q;q^4)_s^2(q;q^4)_{(n-1)/4+s} (q^2;q^4)_{(n-1)/4-s}} {(q^4;q^4)_s^2(q^4;q^4)_{(n-1)/4+s}(q;q^4)_{(n-1)/4-s}} q^{3s+(1-n)/4} \pmod{\Phi_n(q)^3}. \end{align} $$

Here we need to be familiar with the standard q-notation. The q-shifted factorial is defined by $(a;q)_0=1$ and $(a;q)_n=(1-a)(1-aq)\cdots (1-aq^{n-1})$ for any positive integer n. The q-integer is defined as $[n]=(1-q^n)/(1-q)$ and $\Phi _n(q)$ denotes the nth cyclotomic polynomial, which can be written as

$$ \begin{align*} \Phi_n(q)=\prod_{\substack{1\leqslant k\leqslant n\\ \gcd(k,n)=1}}(q-\zeta^k), \end{align*} $$

where $\zeta $ is a primitive nth root of unity.

It is easy to see that (1.3) follows from (1.4) by taking $n=p$ and $q\to 1$ . The $s=0$ case of (1.4) was given by Liu and Wang [9] and can also be deduced from [5, Theorem 4.3].

Swisher [11, (3)] gave the following extension of Van Hamme’s (G.2) supercongruence: for $p\equiv 3 \pmod 4$ and $p>3$ ,

(1.5) $$ \begin{align} &\sum_{k=0}^{(3p-1)/4}(8k+1)\frac{(\frac{1}{4})_k^4}{k!^4} \equiv -\frac{3p^2\Gamma_p(\frac 12)\Gamma_p(\frac 14)}{2\Gamma_p(\frac 34)} \pmod{p^4}. \end{align} $$

(The negative sign was missing in Swisher’s original supercongruence.)

We shall give a new generalisation of (1.5) modulo $p^3$ as follows: for $p\equiv 3\pmod {4}$ and $0\leqslant s\leqslant (p-3)/4$ ,

(1.6) $$ \begin{align} \sum_{k=s}^{(3p-1)/4} (8k+1)\frac{(\frac{1}{4})_{k-s}(\frac{1}{4})_{k+s}(\frac{1}{4})_k^2} {(k-s)!(k+s)!k!^2} \equiv (3p+4s)\frac{(\frac{1}{4})_s^2(\frac{1}{4})_{(3p-1)/4+s} (\frac{1}{2})_{(3p-1)/4-s}} {s!^2(1)_{(3p-1)/4+s}(\frac{1}{4})_{(3p-1)/4-s}} \pmod{p^3}. \end{align} $$

When $s=0$ , the right-hand side of (1.6) reduces to $3p(\tfrac 12)_{(3p-1)/4}/(1)_{(3p-1)/4}$ , which is easily seen to be congruent to the right-hand side of (1.5) modulo $p^3$ . Thus, the supercongruence (1.6) can be deemed a generalisation of the modulo $p^3$ case of (1.5). A result of Guo and Schlosser [5, Corollary 1.2 with $d = 4$ and $q\to 1$ ] implies that (1.6) is even true modulo $p^4$ for $s=0$ . However, numerical evaluation indicates that (1.6) is not true modulo $p^4$ for general s.

In the same way as before, we shall prove (1.6) via the following q-supercongruence.

Theorem 1.2. Let $n\equiv 3\pmod 4$ be a positive integer and let $0\leqslant s\leqslant (n-3)/4$ . Then

(1.7) $$ \begin{align} &\sum_{k=s}^{(3n-1)/4} [8k+1]\frac{(q;q^4)_{k-s}(q;q^4)_{k+s}(q;q^4)_k^2} {(q^4;q^4)_{k-s}(q^4;q^4)_{k+s}(q^4;q^4)_k^2} q^{2k} \notag\\[6pt] &\quad\equiv [3n+4s] \frac{(q;q^4)_s^2(q;q^4)_{(3n-1)/4+s} (q^2;q^4)_{(3n-1)/4-s}} {(q^4;q^4)_s^2(q^4;q^4)_{(3n-1)/4+s}(q;q^4)_{(3n-1)/4-s}} q^{3s+(1-3n)/4} \pmod{\Phi_n(q)^3}. \end{align} $$

Our proof of Theorems 1.1 and 1.2 will use the powerful method of ‘creative microscoping’, which was devised by Guo and Zudilin [6].

2 Proof of Theorem 1.1

We require the following easily proved q-congruence, which was first given by Guo and Schlosser [4, Lemma 3].

Lemma 2.1. Let d, m and n be positive integers with $m\leqslant n-1$ and $dm\equiv -1\pmod {n}$ . Then, for $0\leqslant k\leqslant m$ ,

$$ \begin{align*} \frac{(aq;q^d)_{m-k}}{(q^d/a;q^d)_{m-k}} \equiv (-a)^{m-2k}\frac{(aq;q^d)_k}{(q^d/a;q^d)_k} q^{m(dm-d+2)/2+(d-1)k} \pmod{\Phi_n(q)}. \end{align*} $$

Following Gasper and Rahman’s monograph [2], the basic hypergeometric series $_{r+1}\phi _r$ is defined as

$$ \begin{align*}_{r+1}\phi_{r}\bigg[\begin{array}{@{}c@{}} a_1,a_2,\ldots,a_{r+1}\\[6pt] b_1,b_2,\ldots,b_{r} \end{array};q,\, z \bigg] =\sum_{k=0}^{\infty}\frac{(a_1;q)_k(a_2;q)_k\cdots(a_{r+1};q)_k } {(q;q)_k(b_1;q)_k\cdots(b_{r};q)_k}z^k. \end{align*} $$

Jackson’s ${}_6\phi _5$ summation (see [2, Appendix (II.21)]) can be stated as follows:

(2.1) $$ \begin{align} _6\phi_5 \bigg[\begin{array}{@{}c@{}} a,\, qa^{1/2},\, -qa^{1/2},\, b,\, c,\, q^{-n} \\[6pt] a^{1/2},\, -a^{1/2},\, aq/b,\, aq/c,\, aq^{n+1}\end{array};q,\,\frac{aq^{n+1}}{bc} \bigg] =\frac{(aq;q)_n (aq/bc;q)_n}{(aq/b;q)_n(aq/c;q)_n}. \end{align} $$

To prove Theorem 1.1, we first establish the following result.

Theorem 2.2. Let $n\equiv 1\pmod 4$ be an integer greater than $1$ . Let $0\leqslant s\leqslant (n-1)/4$ and let a be an indeterminate. Then, modulo $\Phi _n(q)(1-aq^n)(a-q^n)$ ,

(2.2) $$ \begin{align} &\sum_{k=s}^{(n-1)/4} [8k+1]\frac{(aq;q^4)_k(q/a;q^4)_k (q;q^4)_{k-s}(q;q^4)_{k+s} } {(aq^4;q^4)_k(q^4/a;q^4)_k(q^4;q^4)_{k-s}(q^4;q^4)_{k+s}}q^{2k} \notag\\[6pt] &\quad\equiv [n+4s] \frac{(aq;q^4)_{s}(q/a;q^4)_{s}(q;q^4)_{(n-1)/4+s} (q^2;q^4)_{(n-1)/4-s}} {(aq^4;q^4)_{s}(q^4/a;q^4)_{s}(q^4;q^4)_{(n-1)/4+s}(q;q^4)_{(n-1)/4-s}} q^{3s+(1-n)/4}. \end{align} $$

Proof. For $a=q^{-n}$ or $a=q^{n}$ , the left-hand side of (2.2) is equal to

(2.3) $$ \begin{align} &\sum_{k=s}^{(n-1)/4} [8k+1]\frac{(q^{1-n};q^4)_k(q^{1+n};q^4)_k (q;q^4)_{k-s}(q;q^4)_{k+s} } {(q^{4-n};q^4)_k(q^{4+n};q^4)_k(q^4;q^4)_{k-s}(q^4;q^4)_{k+s}}q^{2k} \notag\\[6pt] &\quad=\sum_{k=0}^{(n-1)/4-s} [8k+8s+1]\frac{(q^{1-n};q^4)_{k+s}(q^{1+n};q^4)_{k+s} (q;q^4)_{k}(q;q^4)_{k+2s} } {(q^{4-n};q^4)_{k+s}(q^{4+n};q^4)_{k+s}(q^4;q^4)_{k}(q^4;q^4)_{k+2s}}q^{2k+2s} \notag\\[6pt] &\quad=[8s+1] \frac{(q^{1-n};q^4)_{s}(q^{1+n};q^4)_{s} (q;q^4)_{2s} } {(q^{4-n};q^4)_{s}(q^{4+n};q^4)_{s}(q^4;q^4)_{2s}} q^{2s}\notag\\[6pt] &\quad\quad\times {}_{6}\phi_{5}\!\left[\begin{array}{@{}cccccccc@{}} q^{1+8s},&\!\! q^{\frac{9}{2}+4s},&\!\! -q^{\frac{9}{2}+4s}, &\!\! q, &\!\! q^{1+n+4s}, &\!\! q^{1-n+4s} \\[8pt] &\!\! q^{\frac{1}{2}+4s}, &\!\! -q^{\frac{1}{2}+4s}, &\!\! q^{4+8s}, &\!\! q^{4-n+4s}, &\!\! q^{4+n+4s} \end{array}\!\!;q^4, q^2 \right]. \end{align} $$

Letting $q\mapsto q^4$ , $a=q^{1+8s}$ , $b=q$ , $c= q^{1+n+4s}$ and $n\mapsto (n-1)/4-s$ in (2.1), one sees that the right-hand side of (2.3) can be simplified as

$$ \begin{align*} &q^{2s}[8s+1] \frac{(q^{1-n};q^4)_{s}(q^{1+n};q^4)_{s} (q;q^4)_{2s} } {(q^{4-n};q^4)_{s}(q^{4+n};q^4)_{s}(q^4;q^4)_{2s}} \frac{(q^{5+8s};q^4)_{(n-1)/4-s} (q^{3-n+4s};q^4)_{(n-1)/4-s}}{(q^{4+8s};q^4)_{(n-1)/4-s}(q^{4-n+4s};q^4)_{(n-1)/4-s}} \\[6pt] &\quad=[n+4s]\frac{(q^{1-n};q^4)_{s}(q^{1+n};q^4)_{s} } {(q^{4-n};q^4)_{s}(q^{4+n};q^4)_{s}} \frac{(q;q^4)_{(n-1)/4+s} (q^{3-n+4s};q^4)_{(n-1)/4-s}}{(q^4;q^4)_{(n-1)/4+s}(q^{4-n+4s};q^4)_{(n-1)/4-s}} q^{2s}\\[6pt] &\quad=[n+4s]\frac{(q^{1-n};q^4)_{s}(q^{1+n};q^4)_{s} } {(q^{4-n};q^4)_{s}(q^{4+n};q^4)_{s}} \frac{(q;q^4)_{(n-1)/4+s} (q^2;q^4)_{(n-1)/4-s}}{(q^4;q^4)_{(n-1)/4+s}(q;q^4)_{(n-1)/4-s}} q^{3s+(1-n)/4}. \end{align*} $$

Thus, we have proved that (2.2) is true modulo $1-aq^n$ and $a-q^n$ .

Since $n\equiv 1\pmod {4}$ , letting $d=4$ and $m=(n-1)/4$ in Lemma 2.1, we obtain

(2.4) $$ \begin{align} \frac{(aq;q^4)_{m-k}}{(q^4/a;q^d)_{m-k}} \equiv (-a)^{m-2k}\frac{(aq;q^4)_k}{(q^4/a;q^d)_k} q^{m(2m-1)+3k} \pmod{\Phi_n(q)} \end{align} $$

for $0\leqslant k\leqslant m$ . Using this q-congruence, we can easily verify the following congruence, for $m=(n-1)/4$ and $s\leqslant k\leqslant m-s$ ,

(2.5) $$ \begin{align} &[8(m-k)+1]\frac{(aq;q^4)_{m-k}(q/a;q^4)_{m-k} (q;q^4)_{m-k-s}(q;q^4)_{m-k+s} } {(aq^4;q^4)_{m-k}(q^4/a;q^4)_{m-k}(q^4;q^4)_{m-k-s}(q^4;q^4)_{m-k+s}} q^{2m-2k} \notag\\[6pt] &\quad\equiv -[8k+1]\frac{(aq;q^4)_k(q/a;q^4)_k (q;q^4)_{k-s}(q;q^4)_{k+s} } {(aq^4;q^4)_k(q^4/a;q^4)_k(q^4;q^4)_{k-s}(q^4;q^4)_{k+s} } q^{2k} \pmod{\Phi_n(q)}. \end{align} $$

Moreover, for $(n-1)/4-s<k\leqslant (n-1)/4$ , the summand indexed k on the left-hand side of (2.2) is congruent to $0$ modulo $\Phi _n(q)$ because $k+s>(n-1)/4$ and $(q;q^4)_{k+s}$ in the numerator incorporates the factor $1-q^n$ . This means that the left-hand side of (2.2) is congruent to $0$ modulo $\Phi _n(q)$ . Since

$$ \begin{align*}[n+4s](q;q^4)_{(n-1)/4+s}=[n](q;q^4)_{(n-1)/4}(q^{n+4};q^4)_s\equiv 0\pmod{\Phi_n(q)}\end{align*} $$

for $n>1$ , we conclude that (2.2) is also true modulo $\Phi _n(q)$ . Noting that the polynomials $1-aq^n$ , $a-q^n$ and $\Phi _n(q)$ are pairwise relatively prime, we complete the proof of the theorem.

Proof of Theorem 1.1.

For $a=1$ , the denominators on both sides of (2.2) are relatively prime to $\Phi _n(q)$ . Moreover, when $a=1$ the polynomial $(1-aq^n)(a-q^n)=(1-q^n)^2$ contains the factor $\Phi _n(q)^2$ . Therefore, putting $a=1$ in (2.2), we obtain the desired q-supercongruence (1.4).

3 Proof of Theorem 1.2

The proof is similar to that of Theorem 1.1. We first establish the following parametric generalisation of Theorem 1.2.

Theorem 3.1. Let $n\equiv 3\pmod 4$ be a positive integer. Let $0\leqslant s\leqslant (n-3)/4$ and let a be an indeterminate. Then, modulo $\Phi _n(q)(1-aq^{3n})(a-q^{3n})$ ,

(3.1) $$ \begin{align} &\sum_{k=s}^{(3n-1)/4} [8k+1]\frac{(aq;q^4)_k(q/a;q^4)_k (q;q^4)_{k-s}(q;q^4)_{k+s} } {(aq^4;q^4)_k(q^4/a;q^4)_k(q^4;q^4)_{k-s}(q^4;q^4)_{k+s}}q^{2k} \notag\\[6pt] &\quad\equiv [n+4s] \frac{(aq;q^4)_{s}(q/a;q^4)_{s}(q;q^4)_{(3n-1)/4+s} (q^2;q^4)_{(3n-1)/4-s}} {(aq^4;q^4)_{s}(q^4/a;q^4)_{s}(q^4;q^4)_{(3n-1)/4+s}(q;q^4)_{(3n-1)/4-s}} q^{3s+(1-3n)/4}. \end{align} $$

Proof. For $a=q^{-3n}$ or $a=q^{3n}$ , the left-hand side of (3.1) is equal to

(3.2) $$ \begin{align} &\sum_{k=s}^{(3n-1)/4} [8k+1]\frac{(q^{1-3n};q^4)_k(q^{1+3n};q^4)_k (q;q^4)_{k-s}(q;q^4)_{k+s} } {(q^{4-3n};q^4)_k(q^{4+3n};q^4)_k(q^4;q^4)_{k-s}(q^4;q^4)_{k+s}}q^{2k} \notag\\[6pt] &\quad=[8s+1] \frac{(q^{1-3n};q^4)_{s}(q^{1+3n};q^4)_{s} (q;q^4)_{2s} } {(q^{4-3n};q^4)_{s}(q^{4+3n};q^4)_{s}(q^4;q^4)_{2s}} q^{2s}\notag\\[6pt] &\quad\quad\times {}_{6}\phi_{5}\!\left[\begin{array}{@{}cccccccc@{}} q^{1+8s},&\!\! q^{\frac{9}{2}+4s},&\!\! -q^{\frac{9}{2}+4s}, &\!\! q, &\!\! q^{1+3n+4s}, &\!\! q^{1-3n+4s} \\[6pt] &\!\! q^{\frac{1}{2}+4s}, &\!\! -q^{\frac{1}{2}+4s}, &\!\! q^{4+8s}, &\!\! q^{4-3n+4s}, &\!\! q^{4+3n+4s} \end{array}\!\!;q^4, q^2 \right]. \end{align} $$

Letting $q\mapsto q^4$ , $a=q^{1+8s}$ , $b=q$ , $c= q^{1+3n+4s}$ and $n\mapsto (3n-1)/4-s$ in (2.1), one sees that the right-hand side of (3.2) may be written as

$$ \begin{align*} &q^{2s}[8s+1] \frac{(q^{1-3n};q^4)_{s}(q^{1+3n};q^4)_{s} (q;q^4)_{2s} } {(q^{4-3n};q^4)_{s}(q^{4+3n};q^4)_{s}(q^4;q^4)_{2s}} \frac{(q^{5+8s};q^4)_{(3n-1)/4-s} (q^{3-3n+4s};q^4)_{(3n-1)/4-s}}{(q^{4+8s};q^4)_{(3n-1)/4-s}(q^{4-3n+4s};q^4)_{(3n-1)/4-s}} \\[6pt] &=[3n+4s]\frac{(q^{1-3n};q^4)_{s}(q^{1+3n};q^4)_{s} } {(q^{4-3n};q^4)_{s}(q^{4+3n};q^4)_{s}} \frac{(q;q^4)_{(3n-1)/4+s} (q^2;q^4)_{(3n-1)/4-s}}{(q^4;q^4)_{(3n-1)/4+s}(q;q^4)_{(3n-1)/4-s}} q^{3s+(1-3n)/4}. \end{align*} $$

This proves that (2.2) holds modulo $1-aq^{3n}$ and $a-q^{3n}$ .

Since $n\equiv 3\pmod {4}$ , letting $d=4$ and $m=(3n-1)/4$ in Lemma 2.1, we again obtain (2.4) for $0\leqslant k\leqslant m$ . Applying this q-congruence, we can check (2.5) for $m=(3n-1)/4$ and $s\leqslant k\leqslant m-s$ . For $(3n-1)/4-s<k\leqslant (3n-1)/4$ , the summand indexed k on the left-hand side of (3.1) is congruent to $0$ modulo $\Phi _n(q)$ because $k+s>(3n-1)/4$ and $(q;q^4)_{k+s}$ has the factor $1-q^{3n}$ . This implies that the left-hand side of (3.1) is congruent to $0$ modulo $\Phi _n(q)$ . Since

$$ \begin{align*}[3n+4s](q;q^4)_{(3n-1)/4+s}=[3n](q;q^4)_{(3n-1)/4}(q^{3n+4};q^4)_s\equiv 0\pmod{\Phi_n(q)},\end{align*} $$

we conclude that (3.1) is also true modulo $\Phi _n(q)$ .

Proof of Theorem 1.1.

When $a=1$ , the polynomial $(1-aq^{3n})(a-q^{3n})=(1-q^{3n})^2$ contains the factor $\Phi _n(q)^2$ . Thus, letting $a=1$ in (2.2), we get the q-supercongruence (1.7).

4 An open problem

We believe that the following conjecture is true.

Conjecture 4.1. The q-supercongruences (1.4) and (1.7) are also true modulo $[n]\Phi _n(q)^2$ .

The above conjecture is clearly true for $s=0$ (see [5, 9]). Numerical computation indicates that both sides of (1.4) (or (1.7)) should be congruent to $0$ modulo $[n]$ . However, it seems difficult to confirm this. The technique of proving a q-congruence modulo $[n]$ introduced in [6] does not work here, because of the additional parameter s.