On the divisibility of some truncated hypergeometric series

Abstract

Let $p$ be an odd prime and $r\ge 1$. Suppose that $\alpha$ is a $p$-adic integer with $\alpha \equiv 2a(\mathrm{mod}p)$ for some $1\le a<(p+r)/(2r+1)$. We confirm a conjecture of Sun and prove that $${}_{2r+1}{F}_{2r}\left[\begin{array}{cccc}\alpha & \alpha & \dots & \alpha \\ & 1& \dots & 1\end{array}\right|1{]}_{p-1}\equiv 0(\mathrm{mod}{p}^2),$$ where the truncated hypergeometric series $${}_{q+1}{F}_q\left[\begin{array}{cccc}{x}_0& x_1& \dots & x_q\\ & y_1& \dots & y_q\end{array}\right|z{]}_n:=\sum _{k=0}^n\frac{(x_0{)}_k(x_1{)}_k\cdots (x_q{)}_k}{(y_1{)}_k\cdot (y_q{)}_k}\cdot \frac{z^k}{k!}.$$

Theorem 1.2 in the first version has been removed, which is incorrect

References in corpus (2)

• A modular supercongruence for ${}_6{F}_5$: an Apéry-like story
• $p$-adic analogues of hypergeometric identities

Cited by in corpus (1)

• A supercongruence involving cubes of Catalan numbers