p-adic congruences motivated by series

Zhi-Wei Sun

paper

p-adic congruences motivated by series

arXiv:1111.4988

Abstract

Let $p > 5$ be a prime. Motivated by the known formulae $\sum_{k = 1}^{\infty} (- 1)^{k} / (k^{3} (k 2 k)) = - 2 ζ (3) /5$ and $\sum_{k = 0}^{\infty} (k 2 k)^{2} / ((2 k + 1) 1 6^{k}) = 4 G / π$ $(w h er e$ G=\sum_{k=0}^\infty(-1)^k/(2k+1)^2 $i s t h e C a t a l an co n s t an t), w es h o w t ha t$ $\sum_{k = 1}^{(p - 1) /2} \frac{( - 1 ) ^{k}}{k ^{3} ( k 2 k )} \equiv - 2 B_{p - 3} (mod p),$ $\sum_{k = (p + 1) /2}^{p - 1} \frac{( k 2 k ) ^{2}}{( 2 k + 1 ) 1 6 ^{k}} \equiv - \frac{7}{4} p^{2} B_{p - 3} (mod p^{3})$ $, an d$ $\sum_{k = 0}^{(p - 3) /2} \frac{( k 2 k ) ^{2}}{( 2 k + 1 ) 1 6 ^{k}} \equiv - 2 q_{p} (2) - p q_{p} (2)^{2} + \frac{5}{12} p^{2} B_{p - 3} (mod p^{3}),$ $w h er e$ B_0,B_1,\ldots $a r e B er n o u l l in u mb er s an d$ q_p(2) $i s t h e F er ma tq u o t i e n t$ (2^{p-1}-1)/p$.

15 pages, final published version

p-adic congruences motivated by series

Abstract

References in corpus (2)