## Super congruences and Euler numbers.(English)Zbl 1256.11011

Summary: Let $$p > 3$$ be a prime. A $$p$$-adic congruence is called a super congruence if it happens to hold modulo some higher power of $$p$$. The topic of super congruences is related to many fields including Gauss and Jacobi sums and hypergeometric series. We prove that \begin{aligned} \sum_{k = 0}^{(p - 1)/2} \frac{\binom{2k}{k}} {2^k} \equiv (-1)^{(p-1)/2} - p^2 E_{p - 3} \pmod{p^3}, \\ \sum_{k = 1}^{(p - 1)/2} \frac{\binom{2k}{k}} {k} \equiv (-1)^{(p+1)/2} \frac{8}{3} pE_{p-3} \pmod{p^2}, \\ \sum_{k = 0}^{(p-1)/2} \frac{\binom{2k}{k}} {16^k} \equiv (-1)^{(p-1)/ 2} + p^2 E_{p-3} \pmod{p^3}, \end{aligned}
where $$E _{0},E _{1},E _{2}, \dots$$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 $$K: = \sum_{k = 1}^\infty \tfrac{k}{3}/k^2$$ (with (-) the Jacobi symbol), two of which are
$\sum_{k = 1}^\infty \frac{(10k-3)8^k}{k^3 \binom{2k}{k}^2 \binom{3k}{k}} = \frac{\pi^2}{2} \quad\text{and}\quad\sum_{k = 1}^\infty \frac{(15k - 4)(-27)^{k-1}} {k^3 \binom{2k}{k}^2 \binom{3k}{k}} = K.$

### MSC:

 11B65 Binomial coefficients; factorials; $$q$$-identities 11A07 Congruences; primitive roots; residue systems 11B68 Bernoulli and Euler numbers and polynomials 33C20 Generalized hypergeometric series, $${}_pF_q$$

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### References:

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