中文

关于某些截断超几何级数的整除性

数论 2018-03-06 v2 组合数学

摘要

pp 为奇素数且 r1r\geq 1。假设 α\alpha 是一个 pp-adic 整数,且对某个 1a<(p+r)/(2r+1)1\leq a<(p+r)/(2r+1)α2a(modp)\alpha\equiv2a\pmod p。我们证实了 Sun 的一个猜想并证明 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}, 其中截断超几何级数 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!}.

关键词

引用

@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}
}

备注

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