Supercongruences via Beukers' method

Zhi-Hong Sun; Dongxi Ye

paper

Supercongruences via Beukers' method

arXiv:2408.09776

Abstract

Recently, using modular forms F. Beukers posed a unified method that can deal with a large number of supercongruences involving binomial coefficients and ApÃ©ry-like numbers. In this paper, we use Beukers' method to prove some conjectures of the first author concerning the congruences for $k = 0 \sum (p - 1) /2 \frac{( k 2 k ) ^{3}}{m ^{k}}, k = 0 \sum p - 1 \frac{( k 2 k ) ^{2} ( 2 k 4 k )}{m ^{k}}, k = 0 \sum p - 1 \frac{( k 2 k ) ( k 3 k ) ( 3 k 6 k )}{m ^{k}}, n = 0 \sum p - 1 \frac{V _{n}}{m ^{n}}, n = 0 \sum p - 1 \frac{T _{n}}{m ^{n}}, n = 0 \sum p - 1 \frac{D _{n}}{m ^{n}}$ and $\sum_{n = 0}^{p - 1} (- 1)^{n} A_{n}$ modulo $p^{3}$ , where $p$ is an odd prime representable by some suitable binary quadratic form, $m$ is an integer not divisible by $p$ , $V_{n} = \sum_{k = 0}^{n} (k 2 k)^{2} (n - k 2 n - 2 k)^{2}$ , $T_{n} = \sum_{k = 0}^{n} (k n)^{2} (n 2 k)^{2}$ , $D_{n} = \sum_{k = 0}^{n} (k n)^{2} (k 2 k) (n - k 2 n - 2 k)$ and $A_{n}$ is the ApÃ©ry number given by $A_{n} = \sum_{k = 0}^{n} (k n)^{2} (k n + k)^{2}$ .

59 pages