# Supercongruences involving Lucas sequences

Research paper by Zhi-Wei Sun

Indexed on: 11 Oct '16Published on: 11 Oct '16Published in: arXiv - Mathematics - Number Theory

#### Abstract

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 $m\in\mathbb Z\setminus\{0\}$ and $\Delta=m(m-4)$, and let $p>3$ be a prime not dividing $m$. For any positive integer $n$, we show that $$\sum_{k=0}^{pn-1} \frac{\binom{2k}k}{m^k} \equiv \left(\frac{\Delta}p\right) \sum_{r=0}^{n-1}\frac{\binom{2r}r}{m^r} + \frac{n}{m^{n-1}} \binom{2n-1}{n-1} u_{p-(\frac{\Delta}p)}(m-2,1) \pmod{p^2}.$$ In particular, for any prime $p>3$ and positive integer $n$, we have $$\sum_{k=0}^{pn-1} \frac{\binom{2k}k}{2^k} \equiv \left(\frac{-1}p\right) \sum_{r=0}^{n-1}\frac{\binom{2r}r}{2^r}\pmod{p^2} \ \ \mbox{and}\ \ \sum_{k=0}^{pn-1} \frac{\binom{2k}k}{3^k} \equiv \left(\frac{p}3\right) \sum_{r=0}^{n-1} \frac{\binom{2r}r}{3^r} \pmod{p^2}.$$ We also pose some conjectures for further research.