On divisibility of sums of Apery polynomials
arXiv:1108.1546
Abstract
For any positive integers $m$ and $α$, we prove that $$\sum_{k=0}^{n-1}ε^k(2k+1)A_k^{(α)}(x)^m\equiv0\pmod{n}, $$ where $ε\in\{1,-1\}$ and $$ A_n^{(α)}(x)=\sum_{k=0}^n\binom{n}{k}^α\binom{n+k}{k}^αx^k.$$
This is a preliminary draft