English

Supercongruences via Beukers' method

Number Theory 2024-09-20 v4 Classical Analysis and ODEs Combinatorics

Abstract

Recently, using modular forms F. Beukers posed a unified method that can deal with a large number of supercongruences involving binomial coefficients and Ap\'ery-like numbers. In this paper, we use Beukers' method to prove some conjectures of the first author concerning the congruences for k=0(p1)/2(2kk)3mk, k=0p1(2kk)2(4k2k)mk, k=0p1(2kk)(3kk)(6k3k)mk, n=0p1Vnmn, n=0p1Tnmn, n=0p1Dnmn\sum_{k=0}^{(p-1)/2}\frac{\binom{2k}k^3}{m^k}, \ \sum_{k=0}^{p-1}\frac{\binom{2k}k^2\binom{4k}{2k}}{m^k}, \ \sum_{k=0}^{p-1}\frac{\binom{2k}k\binom{3k}k\binom{6k}{3k}}{m^k}, \ \sum_{n=0}^{p-1}\frac{V_n}{m^n},\ \sum_{n=0}^{p-1}\frac{T_n}{m^n},\ \sum_{n=0}^{p-1}\frac{D_n}{m^n} and n=0p1(1)nAn\sum_{n=0}^{p-1}(-1)^nA_n modulo p3p^3, where pp is an odd prime representable by some suitable binary quadratic form, mm is an integer not divisible by pp, Vn=k=0n(2kk)2(2n2knk)2V_n=\sum_{k=0}^n\binom{2k}k^2\binom{2n-2k}{n-k}^2, Tn=k=0n(nk)2(2kn)2T_n=\sum_{k=0}^n\binom nk^2\binom{2k}n^2, Dn=k=0n(nk)2(2kk)(2n2knk)D_n=\sum_{k=0}^n\binom nk^2\binom{2k}k\binom{2n-2k}{n-k} and AnA_n is the Ap\'ery number given by An=k=0n(nk)2(n+kk)2A_n=\sum_{k=0}^n\binom nk^2\binom{n+k}k^2.

Keywords

Cite

@article{arxiv.2408.09776,
  title  = {Supercongruences via Beukers' method},
  author = {Zhi-Hong Sun and Dongxi Ye},
  journal= {arXiv preprint arXiv:2408.09776},
  year   = {2024}
}

Comments

59 pages

R2 v1 2026-06-28T18:16:25.295Z