Yuanbo Yu

In this paper, we establish two $q$-supercongruences for double series with parameters. When the parameters are specified, they reduce to two known results and can produce a lot of new supercongruences including the following two ones: \begin{align*} &\sum _{k=0}^{(p-1)/2}(4k+1)\frac {(\frac {1}{2})_k^3}{(1)_k^3} \sum _{j=1}^{2k}\frac {(-1)^{k+j+1}}{j^2}\equiv p\!\!\sum _{k=0}^{(p-1)/2}\frac {(\frac {1}{2})_k^2}{(1)_k^2} \!\!\sum _{j=1}^{k}\frac {1}{(2j-1)^2} \!\!\!\pmod {p^2}, \end{align*} where $p$ is an odd prime; \begin{align*} &\sum _{k=0}^{(p+1)/2}(-1)^k(4k-1)\frac {(-\frac {1}{2})_k^3}{(1)_k^3} \sum _{j=1}^{k}\bigg \{\frac {1}{(2j)^2}-\frac {1}{(2j-3)^2}\bigg \} \\&\quad \equiv \frac {p}{8}\sum _{k=0}^{(p-3)/2}\frac {(\frac {3}{2})_k^2}{(1)_k(3)_k} \sum _{j=0}^{k}\frac {1}{(2j+1)^2} \pmod {p^2}, \end{align*} where $p>3$ is any prime. It should be pointed out that the two results are both supercongruences for truncated $\pi$-series.