Guillera–Zudilin-type supercongruence for a fifth-power hypergeometric sum

From papers

Let pp be an odd prime, let rr be a positive integer, and let δ{1,2}\delta\in\{1,2\}. Guillera–Zudilin-type conjecture.

n=0(pr1)/δ(12)n5n!5(10n2+6n+1)(4)np2r(modp2r+3).\sum_{n=0}^{(p^r-1)/\delta}\frac{\left(\frac12\right)^5_n}{n!^5}(10n^2+6n+1)(-4)^n\equiv p^{2r}\pmod{p^{2r+3}}.

The paper states that this stronger modulus cannot be proved by its method; the surrounding theorem establishes a weaker, piecewise congruence modulo pr+4p^{r+4}.

Progress summary

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Sources & referencesView supporting material

Primary source

Guo-Shuai Mao, “Proof of some supercongruences via the Wilf-Zeilberger method”, arXiv:1909.13173 (2019).

Solutions 0

No solutions have been posted yet.