A q-congruence implying the Beukers–Van Hamme congruence

IF 1.3 3区数学 Q2 MATHEMATICS, APPLIED

Bulletin des Sciences Mathematiques Pub Date : 2025-03-21 DOI:10.1016/j.bulsci.2025.103615

Victor J.W. Guo , Ji-Cai Liu

引用次数: 0

Abstract

By making use of Andrews' terminating q-analogue of Watson's formula and a double sum identity, we give a q-analogue of the following congruence: for any prime

p \equiv 1 (\mod 4)

\sum_{k = 0}^{(p - 1) / 2} (\begin{matrix} (p - 1) / 2 \\ k \end{matrix}) (\begin{matrix} (p - 1) / 2 + k \\ k \end{matrix}) \equiv \frac{1}{2^{(p - 1) / 2}} (\begin{matrix} (p - 1) / 2 \\ (p - 1) / 4 \end{matrix}) (\mod p^{2}) .

In view of the Chowla–Dwork–Evans congruence, our q-congruence may somewhat be regarded as a q-analogue of the Beukers–Van Hamme congruence:

\sum_{k = 0}^{(p - 1) / 2} (\begin{matrix} (p - 1) / 2 \\ k \end{matrix}) (\begin{matrix} (p - 1) / 2 + k \\ k \end{matrix}) \equiv {(- 1)}^{(p - 1) / 4} (2 a - \frac{p}{2 a}) (\mod p^{2}),

where

p = a^{2} + b^{2}

with

a, b \in Z

and

a \equiv 1 (\mod 4)

查看原文本刊更多论文

q同余意味着Beukers-Van Hamme同余

利用Andrews对Watson公式的终止q-类比和二重和恒等式，我们给出了以下同余的q-类比：对于任意素数p≡1(mod4),∑k=0(p−1)/2((p−1)/2k)((p−1)/2+kk)≡12(p−1)/2((p−1)/2(p−1)/4)（modp2）。鉴于Chowla-Dwork-Evans同余，我们的q-同余在某种程度上可以看作是beukes - van Hamme同余的q-类比：∑k=0(p−1)/2((p−1)/2k)((p−1)/2+kk)≡(−1)(p−1)/4(2a−p2a)(modp2)，其中p=a2+b2, a,b∈Z, a≡1（mod4）。

本文章由计算机程序翻译，如有差异，请以英文原文为准。

求助全文

约1分钟内获得全文求助全文

来源期刊