English

On the divisibility of some truncated hypergeometric series

Number Theory 2018-03-06 v2 Combinatorics

Abstract

Let pp be an odd prime and r1r\geq 1. Suppose that α\alpha is a pp-adic integer with α2a(modp)\alpha\equiv2a\pmod p for some 1a<(p+r)/(2r+1)1\leq a<(p+r)/(2r+1). We confirm a conjecture of Sun and prove that 2r+1F2r[ααα111]p10(modp2),{}_{2r+1}F_{2r}\bigg[\begin{matrix}\alpha&\alpha&\ldots&\alpha\\ &1&\ldots&1\end{matrix}\bigg|\,1\bigg]_{p-1}\equiv0\pmod{p^2}, where the truncated hypergeometric series q+1Fq[x0x1xqy1yqz]n:=k=0n(x0)k(x1)k(xq)k(y1)k(yq)kzkk!. {}_{q+1}F_{q}\bigg[\begin{matrix}x_0&x_1&\ldots&x_{q}\\ &y_1&\ldots&y_q\end{matrix}\bigg|\,z\bigg]_{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!}.

Keywords

Cite

@article{arxiv.1801.02213,
  title  = {On the divisibility of some truncated hypergeometric series},
  author = {Guo-Shuai Mao and Hao Pan},
  journal= {arXiv preprint arXiv:1801.02213},
  year   = {2018}
}

Comments

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