Guo–Schlosser's supercongruence for primes congruent to 2 modulo 3

Let (x)0=1(x)_0=1 and (x)k=x(x+1)(x+k1)(x)_k=x(x+1)\cdots(x+k-1) denote the Pochhammer symbol. Guo–Schlosser's supercongruence. For every prime p2(mod3)p\equiv 2\pmod{3}, one has

k=0(p+1)/3(6k1)(13)k4(1)2k(1)k4(23)2kp(modp3).\sum_{k=0}^{(p+1)/3}(6k-1)\frac{\left(-\frac{1}{3}\right)_k^4(1)_{2k}}{(1)_k^4\left(-\frac{2}{3}\right)_{2k}}\equiv p\pmod{p^3}.

This is a supercongruence for a truncated hypergeometric series. It was proposed by Guo and Schlosser and later confirmed by Jana and Kalita, so it is solved.

Sources & referencesView supporting material

Primary source

Ji-Cai Liu, “Some supercongruences arising from symbolic summation”, arXiv:1912.00663 (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.