Divisibility results concerning truncated hypergeometric series

Abstract

In this paper, using the well-known Karlsson-Minton formula, we mainly establish two divisibility results concerning truncated hypergeometric series. Let $n>2$ and $q>0$ be integers with $2\mid n$ or $2\nmid q$. We show that $$\sum _{k=0}^{p-1}\frac{(q-\frac{p}{n}{)}_k^n}{(1{)}_k^n}\equiv 0(\mathrm{mod}{p}^3)$$ and $$p^n\sum _{k=0}^{p-1}\frac{(1{)}_k^n}{(\frac{p}{n}-q+2{)}_k^n}\equiv 0(\mathrm{mod}{p}^3)$$ for any prime $p>\max \{n,(q-1)n+1\}$, where $(x{)}_k$ denotes the Pochhammer symbol defined by $$(x{)}_k=\{\begin{array}{ll}1,& k=0,\\ x(x+1)\cdots (x+k-1),& k>0.\end{array}$$ Let $n\ge 4$ be an even integer. Then for any prime $p$ with $p\equiv -1(\mathrm{mod}n)$, the first congruence above implies that $$\sum _{k=0}^{p-1}\frac{(\frac{1}{n}{)}_k^n}{(1{)}_k^n}\equiv 0(\mathrm{mod}{p}^3).$$ This confirms a recent conjecture of Guo.

11 pages. Some typos have been corrected