A supercongruence for a truncated hypergeometric series with parameter r

Let rr be a positive integer and let pp be a prime with p>2r+1p>2r+1. The supercongruence.

k=0p1kr(k+1p+1)r(1p+1)kp+1k!p+10(modp4).\sum_{k=0}^{p-1}k^r\left(k+\frac{1}{p+1}\right)^r\frac{\left(\frac{1}{p+1}\right)_k^{p+1}}{k!^{p+1}}\equiv 0\pmod{p^4}.

This is one of three proposed strengthenings related to the paper's earlier supercongruences; its status is conjectural.

Sources & referencesView supporting material

Primary source

Victor J. W. Guo and Michael J. Schlosser, “Some new q-congruences for truncated basic hypergeometric series: even powers”, arXiv:1904.00490 (2019).

Progress summary

Never refreshed

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

Solutions 0

No solutions have been posted yet.