A higher-power supercongruence for a truncated hypergeometric series

From papers

Let p>3p>3 be a prime, and let rr be a positive integer. The supercongruence. First,

k=0p1(2pk+2k+1)(1p+1)k2p+2k!2p+20(modp7).\sum_{k=0}^{p-1}(2pk+2k+1)\frac{\left(\frac{1}{p+1}\right)_k^{2p+2}}{k!^{2p+2}}\equiv 0\pmod{p^7}.

More generally,

k=0pr1(2kpr+11pr1+1)(pr1pr+11)k2pr+11p1k!2pr+11p10(modp2r+5).\sum_{k=0}^{p^r-1}\left(2k\frac{p^{r+1}-1}{p^r-1}+1\right)\frac{\left(\frac{p^r-1}{p^{r+1}-1}\right)_k^{2\frac{p^{r+1}-1}{p-1}}}{k!^{2\frac{p^{r+1}-1}{p-1}}}\equiv 0\pmod{p^{2r+5}}.

This conjecture concerns supercongruences modulo high prime powers and generalizes the displayed case; its status is conjectural.

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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).

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