# Congruences involving $$g_n(x)=\sum \limits _{k=0}^n\left( {\begin{array}{c}n\\ k\end{array}}\right) ^2\left( {\begin{array}{c}2k\\ k\end{array}}\right) x^k$$ g n ( x ) = ∑ k = 0 n n k 2 2 k k x k

Research paper by Zhi-;Wei Sun

Indexed on: 12 Aug '16Published on: 01 Aug '16Published in: The Ramanujan Journal

#### Abstract

Abstract Define $$g_n(x)=\sum _{k=0}^n\left( {\begin{array}{c}n\\ k\end{array}}\right) ^2\left( {\begin{array}{c}2k\\ k\end{array}}\right) x^k$$ for $$n=0,1,2,\ldots$$ . Those numbers $$g_n=g_n(1)$$ are closely related to Apéry numbers and Franel numbers. In this paper we establish some fundamental congruences involving $$g_n(x)$$ . For example, for any prime $$p>5$$ we have \begin{aligned} \sum _{k=1}^{p-1}\frac{g_k(-1)}{k}\equiv 0\pmod {p^2}\quad \text {and}\quad \sum _{k=1}^{p-1}\frac{g_k(-1)}{k^2}\equiv 0\pmod p. \end{aligned} This is similar to Wolstenholme’s classical congruences \begin{aligned} \sum _{k=1}^{p-1}\frac{1}{k}\equiv 0\pmod {p^2}\quad \text {and}\quad \sum _{k=1}^{p-1}\frac{1}{k^2}\equiv 0\pmod p \end{aligned} for any prime $$p>3$$ .AbstractDefine $$g_n(x)=\sum _{k=0}^n\left( {\begin{array}{c}n\\ k\end{array}}\right) ^2\left( {\begin{array}{c}2k\\ k\end{array}}\right) x^k$$ for $$n=0,1,2,\ldots$$ . Those numbers $$g_n=g_n(1)$$ are closely related to Apéry numbers and Franel numbers. In this paper we establish some fundamental congruences involving $$g_n(x)$$ . For example, for any prime $$p>5$$ we have \begin{aligned} \sum _{k=1}^{p-1}\frac{g_k(-1)}{k}\equiv 0\pmod {p^2}\quad \text {and}\quad \sum _{k=1}^{p-1}\frac{g_k(-1)}{k^2}\equiv 0\pmod p. \end{aligned} This is similar to Wolstenholme’s classical congruences \begin{aligned} \sum _{k=1}^{p-1}\frac{1}{k}\equiv 0\pmod {p^2}\quad \text {and}\quad \sum _{k=1}^{p-1}\frac{1}{k^2}\equiv 0\pmod p \end{aligned} for any prime $$p>3$$ . $$g_n(x)=\sum _{k=0}^n\left( {\begin{array}{c}n\\ k\end{array}}\right) ^2\left( {\begin{array}{c}2k\\ k\end{array}}\right) x^k$$ $$g_n(x)=\sum _{k=0}^n\left( {\begin{array}{c}n\\ k\end{array}}\right) ^2\left( {\begin{array}{c}2k\\ k\end{array}}\right) x^k$$ $$n=0,1,2,\ldots$$ $$n=0,1,2,\ldots$$ $$g_n=g_n(1)$$ $$g_n=g_n(1)$$ $$g_n(x)$$ $$g_n(x)$$ $$p>5$$ $$p>5$$ \begin{aligned} \sum _{k=1}^{p-1}\frac{g_k(-1)}{k}\equiv 0\pmod {p^2}\quad \text {and}\quad \sum _{k=1}^{p-1}\frac{g_k(-1)}{k^2}\equiv 0\pmod p. \end{aligned} \begin{aligned} \sum _{k=1}^{p-1}\frac{g_k(-1)}{k}\equiv 0\pmod {p^2}\quad \text {and}\quad \sum _{k=1}^{p-1}\frac{g_k(-1)}{k^2}\equiv 0\pmod p. \end{aligned} \begin{aligned} \sum _{k=1}^{p-1}\frac{1}{k}\equiv 0\pmod {p^2}\quad \text {and}\quad \sum _{k=1}^{p-1}\frac{1}{k^2}\equiv 0\pmod p \end{aligned} \begin{aligned} \sum _{k=1}^{p-1}\frac{1}{k}\equiv 0\pmod {p^2}\quad \text {and}\quad \sum _{k=1}^{p-1}\frac{1}{k^2}\equiv 0\pmod p \end{aligned} $$p>3$$ $$p>3$$