On divisibility of sums of Apery polynomials
  For any positive integers $m$ and $\alpha$, we prove that $$\sum_{k=0}^{n-1}\epsilon^k(2k+1)A_k^{(\alpha)}(x)^m\equiv0\pmod{n}, $$ where $\epsilon\in\{1,-1\}$ and $$ A_n^{(\alpha)}(x)=\sum_{k=0}^n\binom{n}{k}^{\alpha}\binom{n+k}{k}^{\alpha}x^k.$$ 
Hao Pan
math.NT math.CO