On divisibility of sums of Apery polynomials

Abstract

For any positive integers $m$ and $\alpha$, we prove that $$\sum _{k=0}^{n-1}{\epsilon }^k(2k+1)A_k^{(\alpha )}(x{)}^m\equiv 0(\mathrm{mod}n),$$ where $\epsilon \in \{1,-1\}$ and $$A_n^{(\alpha )}(x)=\sum _{k=0}^n{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}^{\alpha }{\left(\genfrac{}{}{0ex}{}{n+k}{k}\right)}^{\alpha }{x}^k.$$

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