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: $$\sum_{k=0}^{(p-1)/2}(4k+1)\binom{-1/2}k^3\equiv (-1)^{(p-1)/2}p\pmod{p^3}.$$ In this paper we show further that \begin{align*}\sum_{k=0}^{p-1}(4k+1)\binom{-1/2}k^3\equiv &\sum_{k=0}^{(p-1)/2}(4k+1)\binom{-1/2}k^3 \\equiv & (-1)^{(p-1)/2}p+p^3E_{p-3} \pmod{p^4},\end{align*}where $E_0,E_1,E_2,\ldots$ are Euler numbers. We also prove that if $p>3$ then $$\sum_{k=0}^{(p-1)/2}\frac{20k+3}{(-2^{10})^k}\binom{4k}{k,k,k,k}\equiv(-1)^{(p-1)/2}p(2^{p-1}+2-(2^{p-1}-1)^2)\pmod{p^4}.$$