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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 \(p>3\) . \(g_n(x)=\sum _{k=0}^n\left( {\begin{array}{c}n\\ k\end{array}}\right) ^2\left( {\begin{array}{c}2k\\ k\end{array}}\right) x^k\) \(g_n(x)=\sum _{k=0}^n\left( {\begin{array}{c}n\\ k\end{array}}\right) ^2\left( {\begin{array}{c}2k\\ k\end{array}}\right) x^k\) \(n=0,1,2,\ldots \) \(n=0,1,2,\ldots \) \(g_n=g_n(1)\) \(g_n=g_n(1)\) \(g_n(x)\) \(g_n(x)\) \(p>5\) \(p>5\) $$\begin{aligned} \sum _{k=1}^{p-1}\frac{g_k(-1)}{k}\equiv 0\pmod {p^2}\quad \text {and}\quad \sum _{k=1}^{p-1}\frac{g_k(-1)}{k^2}\equiv 0\pmod p. \end{aligned}$$ $$\begin{aligned} \sum _{k=1}^{p-1}\frac{g_k(-1)}{k}\equiv 0\pmod {p^2}\quad \text {and}\quad \sum _{k=1}^{p-1}\frac{g_k(-1)}{k^2}\equiv 0\pmod p. \end{aligned}$$ $$\begin{aligned} \sum _{k=1}^{p-1}\frac{1}{k}\equiv 0\pmod {p^2}\quad \text {and}\quad \sum _{k=1}^{p-1}\frac{1}{k^2}\equiv 0\pmod p \end{aligned}$$ $$\begin{aligned} \sum _{k=1}^{p-1}\frac{1}{k}\equiv 0\pmod {p^2}\quad \text {and}\quad \sum _{k=1}^{p-1}\frac{1}{k^2}\equiv 0\pmod p \end{aligned}$$ \(p>3\) \(p>3\)