Let p be a prime and let x be a p-adic integer. We prove two supercongruences for truncated series of the form $$\begin{align*}\sum_{k=1}^{p-1} \frac{(x)_k}{(1)_k}\cdot \frac{1}{k}\sum_{1\le j_1\le\cdots\le j_r\le k}\frac{1}{j_1^{}\cdots j_r^{}}\quad\mbox{and}\quad \sum_{k=1}^{p-1} \frac{(x)_k(1-x)_k}{(1)_k^2}\cdot \frac{1}{k}\sum_{1\le j_1\le\cdots\le j_r\le k}\frac{1}{j_1^{2}\cdots j_r^{2}}\end{align*}$$ which generalise previous results. We also establish q-analogues of two binomial identities.

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与多个谐波和相关的两个超同余

让p是一个素数，让X做一个p-adic 整数。我们证明了以下形式的截断系列的两个超同余$$\begin{align*}\sum_{k=1}^{p-1} \frac{(x)_k}{(1)_k}\cdot \frac{1}{k}\sum_{1\ le j_1\le\cdots\le j_r\le k}\frac{1}{j_1^{}\cdots j_r^{}}\quad\mbox{and}\quad \sum_{k=1}^{p- 1} \frac{(x)_k(1-x)_k}{(1)_k^2}\cdot \frac{1}{k}\sum_{1\le j_1\le\cdots\le j_r\le k}\frac{1}{j_1^{2}\cdots j_r^{2}}\end{align*}$$它概括了以前的结果。我们还建立q- 两个二项式恒等式的类比。