{
  "id": "1610.03384",
  "version": "v1",
  "published": "2016-10-11T18:17:12.000Z",
  "updated": "2016-10-11T18:17:12.000Z",
  "title": "Supercongruences involving Lucas sequences",
  "authors": [
    "Zhi-Wei Sun"
  ],
  "comment": "13 pages",
  "categories": [
    "math.NT"
  ],
  "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.",
  "revisions": [
    {
      "version": "v1",
      "updated": "2016-10-11T18:17:12.000Z"
    }
  ],
  "analyses": {
    "subjects": [
      "11A07",
      "11B39",
      "11B65",
      "05A10",
      "11B75"
    ],
    "keywords": [
      "lucas sequence",
      "positive integer",
      "supercongruences",
      "legendre symbol"
    ],
    "note": {
      "typesetting": "TeX",
      "pages": 13,
      "language": "en",
      "license": "arXiv",
      "status": "editable"
    }
  }
}