Swisher confirmed an interesting congruence: for any odd prime p, $$\begin{aligned} \sum _{k=0}^{(p-1)/2}(-1)^k(6k+1)\frac{(\frac{1}{2})_k^{3}}{k!^{3}8^k} \sum _{j=1}^{k}\Bigg (\frac{1}{(2j-1)^{2}}-\frac{1}{16j^{2}}\Bigg ) \equiv 0 \pmod {p}, \end{aligned}$$ which was conjectured by Long. Recently, its q-analogue was proved by Gu and Guo. Inspired by their work, we obtain a new similar q-congruence modulo $$\Phi _n(q)$$ and two q-supercongruences modulo $$\Phi _n(q)^{2}$$ on double basic hypergeometric sums, where $$\Phi _n(q)$$ is the n-th cyclotomic polynomial.

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关于双和的一些新的 q-congruences

Swisher 证实了一个有趣的同余：对于任何奇素数 p，$$\begin{aligned} \sum _{k=0}^{(p-1)/2}(-1)^k(6k+1)\frac {(\frac{1}{2})_k^{3}}{k!^{3}8^k} \sum _{j=1}^{k}\Bigg (\frac{1}{( 2j-1)^{2}}-\frac{1}{16j^{2}}\Bigg ) \equiv 0 \pmod {p}, \end{aligned}$$ 由 Long 推测。最近，Gu 和 Guo 证明了它的 q 类比。受他们工作的启发，我们在双基本超几何和上获得了一个新的类似 q-congruence modulo $$\Phi _n(q)$$ 和两个 q-supercongruence modulo $$\Phi _n(q)^{2}$$，其中 $$\Phi _n(q)$$ 是第 n 个分圆多项式。