English

Divisibility results concerning truncated hypergeometric series

Number Theory 2020-02-25 v2 Combinatorics

Abstract

In this paper, using the well-known Karlsson-Minton formula, we mainly establish two divisibility results concerning truncated hypergeometric series. Let n>2n>2 and q>0q>0 be integers with 2n2\mid n or 2q2\nmid q. We show that k=0p1(qpn)kn(1)kn0(modp3)\sum_{k=0}^{p-1}\frac{(q-\frac{p}{n})_k^n}{(1)_k^n}\equiv0\pmod{p^3} and pnk=0p1(1)kn(pnq+2)kn0(modp3)p^n\sum_{k=0}^{p-1}\frac{(1)_k^n}{(\frac{p}{n}-q+2)_k^n}\equiv0\pmod{p^3} for any prime p>max{n,(q1)n+1}p>\max\{n,(q-1)n+1\}, where (x)k(x)_k denotes the Pochhammer symbol defined by (x)k={1,k=0,x(x+1)(x+k1),k>0. (x)_k=\begin{cases}1,\quad &k=0,\\ x(x+1)\cdots(x+k-1),\quad &k>0.\end{cases} Let n4n\geq4 be an even integer. Then for any prime pp with p1(modn)p\equiv-1\pmod{n}, the first congruence above implies that k=0p1(1n)kn(1)kn0(modp3).\sum_{k=0}^{p-1} \frac{(\frac{1}{n})_k^n}{(1)_k^n}\equiv0\pmod{p^3}. This confirms a recent conjecture of Guo.

Keywords

Cite

@article{arxiv.2002.08814,
  title  = {Divisibility results concerning truncated hypergeometric series},
  author = {Chen Wang and Wei Xia},
  journal= {arXiv preprint arXiv:2002.08814},
  year   = {2020}
}

Comments

11 pages. Some typos have been corrected

R2 v1 2026-06-23T13:48:15.899Z