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Abstract

The $n$th Schröder number is given by $S_n=\sum_{k=0}^{n}{n\choose k}{n+k\choose k}\frac{1}{k+1}.$ Motivated by these numbers, for any positive integer $\alpha$ we introduce the polynomials \begin{equation*} S_n^{(\alpha)}(x)=\sum_{k=0}^{n}\left({n\atop k}\right)^{\alpha}\left({n+k\atop k}\right)^{\alpha}\frac{x^k}{(k+1)^{\alpha}}. \end{equation*} We prove that for any positive integers $r$, $\alpha$, odd prime $p$ and any integer $m$ not divisible by $p$, and for $\varepsilon=\pm 1$, \begin{align*} &\sum_{k=1}^{p-1}{\varepsilon}^k(2k+1)S_k^{(2\alpha-1)}(m)^r\equiv 0 \pmod{p},\\ &\sum_{k=1}^{p-1}{\varepsilon}^k(2k+1)S_k^{(2\alpha)}(m)^r\equiv -2^r \pmod{p}. \end{align*}