p-adic congruences motivated by series
  Let $p>5$ be a prime. Motivated by the known formulae $\sum_{k=1}^\infty(-1)^k/(k^3\binom{2k}{k})=-2\zeta(3)/5$ and $\sum_{k=0}^\infty \binom{2k}{k}^2/((2k+1)16^k)=4G/\pi$$ (where $G=\sum_{k=0}^\infty(-1)^k/(2k+1)^2$ is the Catalan constant), we show that $$\sum_{k=1}^{(p-1)/2}\frac{(-1)^k}{k^3\binom{2k}{k}}\equiv-2B_{p-3}\pmod{p},$$ $$\sum_{k=(p+1)/2}^{p-1}\frac{\binom{2k}{k}^2}{(2k+1)16^k}\equiv-\frac 7{4}p^2B_{p-3}\pmod{p^3}$$, and $$\sum_{k=0}^{(p-3)/2}\frac{\binom{2k}{k}^2}{(2k+1)16^k} \equiv-2q_p(2)-pq_p(2)^2+\frac{5}{12}p^2B_{p-3}\pmod{p^3},$$ where $B_0,B_1,\ldots$ are Bernoulli numbers and $q_p(2)$ is the Fermat quotient $(2^{p-1}-1)/p$. 
Zhi-Wei Sun
math.NT math.CO