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Proof of a q-supercongruence conjectured by Guo and Schlosser
Revista de la Real Academia de Ciencias Exactas, Físicas y Naturales. Serie A. Matemáticas ( IF 1.8 ) Pub Date : 2020-08-19 , DOI: 10.1007/s13398-020-00923-2
Long Li , Su-Dan Wang

In this paper, we confirm the following conjecture of Guo and Schlosser: for any odd integer $n>1$ and $M=(n+1)/2$ or $n-1$, $$ \sum_{k=0}^{M}[4k-1]_{q^2}[4k-1]^2\frac{(q^{-2};q^4)_k^4}{(q^4;q^4)_k^4}q^{4k}\equiv (2q+2q^{-1}-1)[n]_{q^2}^4\pmod{[n]_{q^2}^4\Phi_n(q^2)}, $$ where $[n]=[n]_q=(1-q^n)/(1-q),(a;q)_0=1,(a;q)_k=(1-a)(1-aq)\cdots(1-aq^{k-1})$ for $k\geq 1$ and $\Phi_n(q)$ denotes the $n$-th cyclotomic polynomial.

中文翻译:

郭和施洛瑟猜想的q-超同余证明

在本文中,我们证实了 Guo 和 Schlosser 的以下猜想:对于任何奇整数 $n>1$ 和 $M=(n+1)/2$ 或 $n-1$,$$ \sum_{k=0 }^{M}[4k-1]_{q^2}[4k-1]^2\frac{(q^{-2};q^4)_k^4}{(q^4;q^ 4)_k^4}q^{4k}\equiv (2q+2q^{-1}-1)[n]_{q^2}^4\pmod{[n]_{q^2}^4 \Phi_n(q^2)}, $$ 其中 $[n]=[n]_q=(1-q^n)/(1-q),(a;q)_0=1,(a;q) _k=(1-a)(1-aq)\cdots(1-aq^{k-1})$ 表示 $k\geq 1$ 和 $\Phi_n(q)$ 表示第 $n$ 个分圆多项式.
更新日期:2020-08-19
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