Super congruences and Euler numbers

preprint2011arXivOpen access

Abstract

Let p>3p>3 be a prime. We prove that ∑k=0p−1(2kk)/2k=(−1)(p−1)/2−p2Ep−3(modp3),\sum_{k=0}^{p-1}\binom{2k}{k}/2^k=(-1)^{(p-1)/2}-p^2E_{p-3} (mod p^3), ∑k=1(p−1)/2(2kk)/k=(−1)(p+1)/28/3∗pEp−3(modp2),\sum_{k=1}^{(p-1)/2}\binom{2k}{k}/k=(-1)^{(p+1)/2}8/3*pE_{p-3} (mod p^2), ∑k=0(p−1)/2(2kk)2/16k=(−1)(p−1)/2+p2Ep−3(modp3)\sum_{k=0}^{(p-1)/2}\binom{2k}{k}^2/16^k=(-1)^{(p-1)/2}+p^2E_{p-3} (mod 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 π2π^2, π−2π^{-2} and the constant K:=∑k>0(k/3)/k2K:=\sum_{k>0}(k/3)/k^2 (with (-) the Jacobi symbol), two of which are ∑k=1∞(10k−3)8k/(k3(2kk)2(3kk))=π2/2\sum_{k=1}^\infty(10k-3)8^k/(k^3\binom{2k}{k}^2\binom{3k}{k})=π^2/2 and ∑k>0(15k−4)(−27)k−1/(k3(2kk)2(3kk))=K.\sum_{k>0}(15k-4)(-27)^{k-1}/(k^3\binom{2k}{k}^2\binom{3k}k)=K.

Explore this paper’s authors, topics and related work.

Building this map preview

BZPEER is loading the nearby papers, people, topics and institutions for this page.

Reviews 0

Write a reviewWrite

No reviews yet.

Discussion 0

Add a commentWrite

No comments yet.