1 Introduction
In his first letter to Hardy in 1913, Ramanujan mentioned the following formula (see [Reference Berndt and Rankin1, page 25, (2)]):
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 [Reference Hardy7]. In 1997, Van Hamme [Reference Van Hamme12] conjectured 13 p-adic analogues of Ramanujan’s series, including
(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 [Reference Morita10]. Swisher [Reference Swisher11] and He [Reference He8] 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 $0\leqslant s\leqslant (p-1)/4$ ,
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 [Reference Liu and Wang9]). 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 [Reference Guo3].
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
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
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 [Reference Liu and Wang9] and can also be deduced from [Reference Guo and Schlosser5, Theorem 4.3].
Swisher [Reference Swisher11, (3)] gave the following extension of Van Hamme’s (G.2) supercongruence: for $p\equiv 3 \pmod 4$ and $p>3$ ,
(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 $0\leqslant s\leqslant (p-3)/4$ ,
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 [Reference Guo and Schlosser5, 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
Our proof of Theorems 1.1 and 1.2 will use the powerful method of ‘creative microscoping’, which was devised by Guo and Zudilin [Reference Guo and Zudilin6].
2 Proof of Theorem 1.1
We require the following easily proved q-congruence, which was first given by Guo and Schlosser [Reference Guo and Schlosser4, 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 $0\leqslant k\leqslant m$ ,
Following Gasper and Rahman’s monograph [Reference Gasper and Rahman2], the basic hypergeometric series $_{r+1}\phi _r$ is defined as
Jackson’s ${}_6\phi _5$ summation (see [Reference Gasper and Rahman2, Appendix (II.21)]) can be stated as follows:
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 $\Phi _n(q)(1-aq^n)(a-q^n)$ ,
Proof. For $a=q^{-n}$ or $a=q^{n}$ , the left-hand side of (2.2) is equal to
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
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
for $0\leqslant k\leqslant m$ . Using this q-congruence, we can easily verify the following congruence, for $m=(n-1)/4$ and $s\leqslant k\leqslant m-s$ ,
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
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.
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 $\Phi _n(q)(1-aq^{3n})(a-q^{3n})$ ,
Proof. For $a=q^{-3n}$ or $a=q^{3n}$ , the left-hand side of (3.1) is equal to
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
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
we conclude that (3.1) is also true modulo $\Phi _n(q)$ .
4 An open problem
We believe that the following conjecture is true.
The above conjecture is clearly true for $s=0$ (see [Reference Guo and Schlosser5, Reference Liu and Wang9]). 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 [Reference Guo and Zudilin6] does not work here, because of the additional parameter s.