Super congruences and Euler numbers

Zhi-Wei Sun

doi:10.1007/s11425-011-4302-x

paper

Super congruences and Euler numbers

arXiv:1001.4453 · doi:10.1007/s11425-011-4302-x

Abstract

Let $p > 3$ be a prime. We prove that $k = 0 \sum p - 1 (k 2 k) / 2^{k} = (- 1)^{(p - 1) /2} - p^{2} E_{p - 3} (m o d p^{3}),$ $k = 1 \sum (p - 1) /2 (k 2 k) / k = (- 1)^{(p + 1) /2} 8/3 * p E_{p - 3} (m o d p^{2}),$ $k = 0 \sum (p - 1) /2 (k 2 k)^{2} /1 6^{k} = (- 1)^{(p - 1) /2} + p^{2} E_{p - 3} (m o d p^{3})$ , where E_0,E_1,E_2,... are Euler numbers. Our new approach is of combinatorial nature. We also formulate many conjectures concerning super congruences and relate most of them to Euler numbers or Bernoulli numbers. Motivated by our investigation of super congruences, we also raise a conjecture on 7 new series for $π^{2}$ , $π^{- 2}$ and the constant $K := \sum_{k > 0} (k /3) / k^{2}$ (with (-) the Jacobi symbol), two of which are $k = 1 \sum \infty (10 k - 3) 8^{k} / (k^{3} (k 2 k)^{2} (k 3 k)) = π^{2} /2$ and $k > 0 \sum (15 k - 4) (- 27)^{k - 1} / (k^{3} (k 2 k)^{2} (k 3 k)) = K .$

References in corpus (1)

Binomial coefficients, Catalan numbers and Lucas quotients

Super congruences and Euler numbers

Abstract

References in corpus (1)

Cited by in corpus (16)