How to prove the following two congruences?

Question1: Let $p\equiv 1 \pmod 3$ be a prime, then $$\sum_{k=0\atop k\neq(p-1)/3}^{(p-1)/2}\frac{\binom{2k}k}{3k+1}\equiv 0 \pmod p.$$

Question2: For any odd prime $p>3$, we have
$$\sum_{k=1}^{(p-1)/2}\frac{1}{k\binom{2k}k}\sum_{j=1}^k\frac{\binom{2j}j}{j}\equiv \frac{1}3B_{p-2}\left(\frac{1}3\right) \pmod p,$$ where $B_n(x)$ are the Bernoulli polynomials defined by $$\sum_{n=0}^{\infty}B_n(x)\frac{t^n}{n!}=\frac{te^{xt}}{e^t-1}.$$

  • 1
    $\begingroup$ The first congruence resembles some of the conjectures of van Hamme. $\endgroup$ – EFinat-S Oct 15 '18 at 2:08
  • 1
    $\begingroup$ Without any background provided, the two congruences do not look interesting enough.I guess the poster needs them as lemmas to prove some more interesting congruences. $\endgroup$ – Zhi-Wei Sun Oct 16 '18 at 2:59

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.