Supercongruences via Beukers' method

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

文献摘要

\sum_{k = 0}^{(p - 1) / 2} \frac{{(\binom{2 k}{k})}^{3}}{m^{k}}, \sum_{k = 0}^{p - 1} \frac{{(\binom{2 k}{k})}^{2} (\binom{4 k}{2 k})}{m^{k}}, \sum_{k = 0}^{p - 1} \frac{(\binom{2 k}{k}) (\binom{3 k}{k}) (\binom{6 k}{3 k})}{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}}

Supercongruences via Beukers' method

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Supercongruences via Beukers' method

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Supercongruences via Beukers' method

文献摘要

讨论

|讲解博客

|AI智能校验

Supercongruences via Beukers' method

文献摘要

讨论

|讲解博客

|AI智能校验

讲解博客

AI智能校验

讲解博客

AI智能校验