Note on super congruences modulo $p^2$

Abstract

Let $p$ be an odd prime, and let $m$ be an integer with $p\nmid m$. In this paper show that $$\sum _{k=0}^{p-1}\frac{\left(\genfrac{}{}{0ex}{}{2k}{k}\right)\left(\genfrac{}{}{0ex}{}{a}{k}\right)\left(\genfrac{}{}{0ex}{}{-1-a}{k}\right)}{m^k}\equiv 0(\mathrm{mod}p)implies\sum _{k=0}^{p-1}\frac{\left(\genfrac{}{}{0ex}{}{2k}{k}\right)\left(\genfrac{}{}{0ex}{}{a}{k}\right)\left(\genfrac{}{}{0ex}{}{-1-a}{k}\right)}{m^k}\equiv 0(\mathrm{mod}{p}^2).$$

8 pages

References in corpus (2)

• Open Conjectures on Congruences
• Congruences for Domb and Almkvist-Zudilin numbers