arXiv 1610.033842016-10math.NT

包含Lucas序列的超同余

Supercongruences involving Lucas sequences

Zhi-Wei Sun

备注：: 26 pages, refined version

摘要：

对于$A,B\in\mathbb Z$，Lucas序列$u_n(A,B)\ (n=0,1,2,\ldots)$定义为$u_0(A,B)=0$，$u_1(A,B)=1$，$u_{n+1}(A,B) = Au_n(A,B)-Bu_{n-1}(A,B)$ $(n=1,2,3,\ldots).$.对于任意奇素数$p$和正整数$n$，我们建立了新的结果$$\frac{u_{pn}(A,B) - (\frac{A^2-4B}p) u_n(A,B)}{pn} \in \mathbb Z_p,$$，其中$(\frac{\cdot}p)$是Legendre符号，$\mathbb Z_p$是$p$ -adic整数环.令$p$是一个奇素数，令$n$是一个正整数。对于任何整数$m\not\equiv0\pmod p$，我们表明，$$\frac1{pn}\bigg(\sum_{k=0}^{pn-1} \frac{\binom{2k}k}{m^k} -\left(\fracΔp\right) \sum_{r=0}^{n-1}\frac{\binom{2r}r}{m^r}\bigg)\in\mathbb Z_p$$，而且$$\frac1n\bigg(\sum_{k=0}^{pn-1} \frac{\binom{2k}k}{m^k} -\left(\fracΔp\right) \sum_{r=0}^{n-1}\frac{\binom{2r}r}{m^r}\bigg)\equiv \frac{\binom{2n-1}{n-1}}{m^{n-1}} u_{p-(\fracΔp)}(m-2,1) \pmod{p^2}$$其中$Δ=m(m-4)$。并提出了进一步研究的建议。

英文摘要：

For $A,B\in\mathbb Z$, the Lucas sequence $u_n(A,B)\ (n=0,1,2,\ldots)$ are defined by $u_0(A,B)=0$, $u_1(A,B)=1$, and $u_{n+1}(A,B) = Au_n(A,B)-Bu_{n-1}(A,B)$ $(n=1,2,3,\ldots).$ For any odd prime $p$ and positive integer $n$, we establish the new result $$\frac{u_{pn}(A,B) - (\frac{A^2-4B}p) u_n(A,B)}{pn} \in \mathbb Z_p,$$ where $(\frac{\cdot}p)$ is the Legendre symbol and $\mathbb Z_p$ is the ring of $p$-adic integers. Let $p$ be an odd prime and let $n$ be a positive integer. For any integer $m\not\equiv0\pmod p$, we show that $$\frac1{pn}\bigg(\sum_{k=0}^{pn-1} \frac{\binom{2k}k}{m^k} -\left(\fracΔp\right) \sum_{r=0}^{n-1}\frac{\binom{2r}r}{m^r}\bigg)\in\mathbb Z_p$$ and furthermore $$\frac1n\bigg(\sum_{k=0}^{pn-1} \frac{\binom{2k}k}{m^k} -\left(\fracΔp\right) \sum_{r=0}^{n-1}\frac{\binom{2r}r}{m^r}\bigg)\equiv \frac{\binom{2n-1}{n-1}}{m^{n-1}} u_{p-(\fracΔp)}(m-2,1) \pmod{p^2}$$ where $Δ=m(m-4)$. We also pose some conjectures for further research.

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。

摘要：

英文摘要：

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