q-Analogues of two Ramanujan-type supercongruences

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Abstract

In this paper, we establish a new q-congruence with parameters modulo the fourth power of a cyclotomic polynomial from Watson's ϕ78 transformation. Taking parameters substitution in our new q-congruence, we present a q-analogue of the (D.2) supercongruence of Van Hamme and several different q-analogues of He's supercongruence for p1(mod3). Finally, we propose a Dwork-type supercongruence conjecture on He's supercongruence.

Introduction

In 1997, Van Hamme [27] developed several p-adic analogues of Ramanujan-type formulas for primes p. For example, the infinity seriesk=0(4k+1)(12)k4k!4=,k=0(6k+1)(13)k6k!6=1.01226, have the following nice p-adic analogues(C.2)k=0(p1)/2(4k+1)(12)k4k!4p(modp3)for p>2,(D.2)k=0(p1)/3(6k+1)(13)k6k!6pΓp(1/3)9(modp4)for p1(mod6). Here (a)n=a(a+1)(a+n1) is the Pochhammer symbol and Γp(x) denotes the p-adic Gamma function [23]. Van Hamme [27] himself proved (1.1) and the modulus p2 case of (1.2). In 2011, Long [21] established the following generalization of (1.1):k=0(p1)/2(4k+1)(12)k4k!4p(modp4); and another interesting supercongruence:k=0(p1)/2(4k+1)(12)k6k!6pk=0(p1)/2(12)k4k!4(modp4). Later, Long and Ramakrishna [22, Theorem 2] generalized Van Hamme's conjecture (D.2) to modulus p6 case. They [22, Proposition 25] also gave the following supercongruencesk=0p1(13)k3k!3{Γp(1/3)6(modp3),if p1(mod6);p23Γp(1/3)6(modp3),if p5(mod6). In addition, He [15, Corollary 1.3] obtained the following Ramanujan-type supercongruences: for p5,k=0(p1)/3(6k+1)(13)k4k!4pΓp(1/3)3(modp4),if p1(mod3),k=0(2p1)/3(6k+1)(13)k4k!42pΓp(1/3)3(modp4),if p2(mod3).

During the past few years, finding q-analogues of known congruences or supercongruence has become an important thing for more and more scholars. Especially, most of Van Hamme's supercongruences have been generalized to the q-world with various techniques [2], [3], [4], [5], [6], [7], [8], [10], [11], [12], [13], [30], [31]. Some other recent progress on q-congruences can be found in [9], [14], [16], [17], [18], [19], [20], [24], [25], [28], [29], [32]. For example, Guo and Schlosser [11, Theorem 2.2] proposed the following partial q-analogue of (1.4),k=0(n1)/2[4k+1](q;q2)k6(q2;q2)k6[n]q(1n)/2k=0(n1)/2(q;q2)k4(q2;q2)k4q2k,(mod[n]Φn(q)2), where n is a positive odd integer, and Guo and Wang [12] reproved (1.3) by finding the following q-congruencek=0(n1)/2[4k+1](q;q2)k4(q2;q2)k4[n]q(1n)/2+[n]3q(1n)/2(n21)(1q)224(mod[n]Φn(q)3).

By a result of Guo and Schlosser [10, Theorem 1.1], for n2(mod3), we havek=0(2n1)/3[6k+1](q;q3)k4(q3;q3)k4qkq(12n)/3(q2;q3)(2n1)/3(q3;q3)(2n1)/3[2n](mod[n]Φn(q)3), which is a q-analogue of He's supercongruence (1.7).

Here and in what follows, for an indeterminate q, (a;q)n=(1a)(1aq)(1aqn1) is the q-shifted factorial and (a1,a2,,am;q)n=(a1;q)n(a2;q)n(am;q)n denotes a product of q-shifted factorials, [n]=[n]q=(1qn)/(1q) is the q-integer. Moreover, the n-th cyclotomic polynomial Φn(q) is defined as the following productΦn(q)=1kngcd(n,k)=1(qζk), where ζ is an n-th primitive root of unity.

Inspired by the above work, we shall propose a new q-congruence modulo the fourth power of a cyclotomic polynomial, from which we can deduce a q-analogue of the (D.2) supercongruence of Van Hamme and a q-analogue of He's supercongruence (1.6).

The rest of the paper is organized as follows. We will give our main results in the next section. Then the proofs of our q-congruences will be given in Sections 3 and 4 by combining Guo and Zudilin's ‘creative microscoping’ method with the Chinese remainder theorem for coprime polynomials.

Section snippets

The main results

Our first result can be stated as follows.

Theorem 1

Let n1(mod3) be positive integer. Then, modulo [n]Φn(q)3,k=0M[6k+1](q;q3)k4(cq,dq;q3)k(q3;q3)k4(q3/c,q3/d;q3)k(cd)kq3k(q2;q3)(n1)/3(q3;q3)(n1)/3[n]q(1n)/3{1+[n]2A(n,q)}k=0M(q2/cd;q3)k(q;q3)k3(q3,q3/c,q3/d,q2;q3)kq3k, where M=(n1)/3 or n1, andA(n,q)=(2n+1)(n1)(1q)218+(n+2)(1q)3k=1(n1)/3q3k1[3k1]12(k=1(n1)/3q3k1[3k1])2+12k=1(n1)/3q6k2[3k1]2.

In order to illustrate that our q-congruences are indeed q-analogues of the

Proof of Theorem 1

In order to prove Theorem 1, we need the following results.

Lemma 1

Let d,n be positive integers with gcd(d,n)=1. Let r be an integer and let a, b, c and e be indeterminates. Then, modulo [n],k=0m1[2dk+r](qr,cqr,eqr,qr/b,aqr,qr/a;qd)k(qd,qd/c,qd/e,qdb,qd/a,aqd;qd)k(bce)kq(2d3r)k0;k=0n1[2dk+r](qr,cqr,eqr,qr/b,aqr,qr/a;qd)k(qd,qd/c,qd/e,qdb,qd/a,aqd;qd)k(bce)kq(2d3r)k0, where 0m1n1 and dm1r(modn).

Proof

Let cq(k) denote the k-th term on the left-hand side in (3.1), i.e.,cq(k)=[2dk+r](qr,cqr,eqr,qr/b

Proof of Theorem 6 and Proposition 2

Proof of Theorem 6

At the end of [10], Guo and Schlosser proved that, for positive integers n and d satisfying n1(modd), modulo [n](1aqn)(aqn)(bqn),k=0(n1)/d[2dk+1](aq,q/a,q/b,q;qd)k(aqd,qd/a,bqd,qd;qd)kbkq(d2)k(b/q)(n1)/d(q2/b;qd)(n1)/d(bqd;qd)(n1)/d(bqn)(ab1a2+aqn)(ab)(1ab)[n]+(q,qd1;qd)(n1)/d(aqd,qd/a;qd)(n1)/d[n]. They pointed that, for d3, there is no concrete q-congruence modulo Φn(q)4 from (4.1) by letting b=1 and a1. Nevertheless, taking the specialization d=3 in (4.1), modulo [n](1aq

Conjectures

In 2015, Swisher [26] conjectured a series of Dwork-type congruences about Van Hamme's first 12 supercongruences, such as: for p1(mod3),k=0(pr1)/3(6k+1)(13)k6k!6pΓp(1/3)9k=0(pr11)/3(6k+1)(13)k6k!6(modp6r). According to numerical verification, we think that (5.1) have a companion, ask=0pr1(6k+1)(13)k6k!6pΓp(1/3)9k=0pr11(6k+1)(13)k6k!6(modp6r). Motivated by the above conjecture of Swisher, we put forward the following general congruence conjecture about He's supercongruence (1.6).

Conjecture 1

Acknowledgements

The authors thank the anonymous referees and the editor for many valuable comments on a previous version of this manuscript.

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    This work is supported by National Natural Science Foundation of China (11661032).

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