A refinement of a congruence result by van Hamme and Mortenson

Abstract

Let $p$ be an odd prime. In 2008, E. Mortenson proved van Hamme’s following conjecture:

\[\sum_{k=0}^{(p-1)/2}(4k+1)\Bigl({\matrix{-1/2\\k}}\Bigr)^{3}\equiv(-1)^{(p-1)/2}p\ \bigl(\operatorname{mod}p^{3}\bigr).\]

In this paper, we show further that

\begin{eqnarray*}\sum_{k=0}^{p-1}(4k+1)\Bigl({\matrix{-1/2\\k}}\Bigr)^{3}&\equiv&\sum_{k=0}^{(p-1)/2}(4k+1)\Bigl({\matrix{-1/2\\k}}\Bigr)^{3}\\[-2pt]&\equiv&(-1)^{(p-1)/2}p+p^{3}E_{p-3}\ \bigl(\operatorname{mod}p^{4}\bigr),\end{eqnarray*}

where $E_{0},E_{1},E_{2},\ldots$ are Euler numbers. We also prove that if $p>3$ then

\begin{eqnarray*}&&\sum_{k=0}^{(p-1)/2}\frac{20k+3}{(-2^{10})^{k}}\Bigl({\matrix{4k\\k,k,k,k}}\Bigr)\\&&\quad \equiv(-1)^{(p-1)/2}p\bigl(2^{p-1}+2-\bigl(2^{p-1}-1\bigr)^{2}\bigr)\ \bigl(\operatorname{mod}p^{4}\bigr).\end{eqnarray*}