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arXiv 1504.05482math.NTmath.CO
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关于$q$ -二项式系数幂和的一个同余式的证明

Proof of a congruence on sums of powers of $q$-binomial coefficients

Victor J. W. Guo, Ji-Cai Liu

摘要:

我们证明了,如果$m,n\geqslant 1$和$a_1,\ldots,a_m$是非负整数,那么\begin{align*} \frac{[a_1+\cdots+a_m+1]!}{[a_1]!\ldots[a_m]!}\sum^{n-1}_{h=0}q^h\prod_{i=1}^m{h\brack a_i} \equiv 0\pmod{[n]}, \end{align*},其中$[n]=\frac{1-q^n}{1-q}$,$[n]!=[n][n-1]\cdots[1]$,和${a\brack b}=\prod_{k=1}^b\frac{1-q^{a-k+1}}{1-q^k}$。$a_1=\cdots=a_m$的案例证实了Z. W.太阳我们还证明了,如果$p>\max\{a,b\}$是素数,则\begin{align*} \frac{[a+b+1]!}{[a]![b]!}\sum_{h=0}^{p-1}q^h{h\brack a}{h\brack b} \equiv (-1)^{a-b} q^{ab-{a\choose 2}-{b\choose 2}}[p]\pmod{[p]^2}. \end{align*}

英文摘要:

We prove that, if $m,n\geqslant 1$ and $a_1,\ldots,a_m$ are nonnegative integers, then \begin{align*} \frac{[a_1+\cdots+a_m+1]!}{[a_1]!\ldots[a_m]!}\sum^{n-1}_{h=0}q^h\prod_{i=1}^m{h\brack a_i} \equiv 0\pmod{[n]}, \end{align*} where $[n]=\frac{1-q^n}{1-q}$, $[n]!=[n][n-1]\cdots[1]$, and ${a\brack b}=\prod_{k=1}^b\frac{1-q^{a-k+1}}{1-q^k}$. The $a_1=\cdots=a_m$ case confirms a recent conjecture of Z.-W. Sun. We also show that, if $p>\max\{a,b\}$ is a prime, then \begin{align*} \frac{[a+b+1]!}{[a]![b]!}\sum_{h=0}^{p-1}q^h{h\brack a}{h\brack b} \equiv (-1)^{a-b} q^{ab-{a\choose 2}-{b\choose 2}}[p]\pmod{[p]^2}. \end{align*}

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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