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Super congruences and Euler numbers

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arxiv 1001.4453 v20 pith:PBANZTOU submitted 2010-01-25 math.NT math.CO

Super congruences and Euler numbers

classification math.NT math.CO
keywords binomnumberscongruenceseulersuperapproachbernoullicombinatorial
verification ladder T0 review T1 audit T2 compute T3 formal T4 reserved
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Let $p>3$ be a prime. We prove that $$\sum_{k=0}^{p-1}\binom{2k}{k}/2^k=(-1)^{(p-1)/2}-p^2E_{p-3} (mod p^3),$$ $$\sum_{k=1}^{(p-1)/2}\binom{2k}{k}/k=(-1)^{(p+1)/2}8/3*pE_{p-3} (mod p^2),$$ $$\sum_{k=0}^{(p-1)/2}\binom{2k}{k}^2/16^k=(-1)^{(p-1)/2}+p^2E_{p-3} (mod p^3)$$, 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 $\pi^2$, $\pi^{-2}$ and the constant $K:=\sum_{k>0}(k/3)/k^2$ (with (-) the Jacobi symbol), two of which are $$\sum_{k=1}^\infty(10k-3)8^k/(k^3\binom{2k}{k}^2\binom{3k}{k})=\pi^2/2$$ and $$\sum_{k>0}(15k-4)(-27)^{k-1}/(k^3\binom{2k}{k}^2\binom{3k}k)=K.$$

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