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Proof of a conjecture of Z.-W. Sun on the divisibility of a triple sum

The numbers $R_n$ and $W_n$ are defined as \begin{align*} R_n=\sum_{k=0}^{n}{n+k\choose 2k}{2k\choose k}\frac{1}{2k-1},\ \text{and}\ W_n=\sum_{k=0}^{n}{n+k\choose 2k}{2k\choose k}\frac{3}{2k-3}. \end{align*} We prove that, for any positive integer $n$ and odd prime $p$, there hold \begin{align*} \sum_{k=0}^{n-1}(2k+1)R_k^2 &\equiv 0 \pmod{n}, \\ \sum_{k=0}^{p-1}(2k+1)R_k^2 &\equiv 4p(-1)^{\frac{p-1}{2}} -p^2 \pmod{p^3}, \\ 9\sum_{k=0}^{n-1}(2k+1)W_k^2 &\equiv 0 \pmod{n}, \\ \sum_{k=0}^{p-1}(2k+1)W_k^2 &\equiv 12p(-1)^{\frac{p-1}{2}}-17p^2 \pmod{p^3}, \quad\text{if $p>3$.} \end{align*} The first two congruences were originally conjectured by Z.-W. Sun. Our proof is based on the multi-variable Zeilberger algorithm and the following observation: $$ {2n\choose n}{n\choose k}{m\choose k}{k\choose m-n}\equiv 0\pmod{{2k\choose k}{2m-2k\choose m-k}}, $$ where $0\leqslant k\leqslant n\leqslant m \leqslant 2n$.

preprint2015arXivOpen access

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