{
  "id": "1511.06221",
  "version": "v1",
  "published": "2015-11-17T01:36:25.000Z",
  "updated": "2015-11-17T01:36:25.000Z",
  "title": "Proof of some congruences conjectured by Guo and Liu",
  "authors": [
    "Guo-Shuai Mao"
  ],
  "comment": "14 pages, 0 figures",
  "categories": [
    "math.NT",
    "math.CO"
  ],
  "abstract": "Let $n$ and $r$ be positive integers and $p$ a prime. Define the numbers $S_n^{(r)}$ and $T_n^{(r)}$ by $$S_n^{(r)}=\\sum_{k=0}^n\\binom{n}{k}^2\\binom{2k}{k}(2k+1)^r\\ \\mbox{,}\\ T_n^{(r)}=\\sum_{k=0}^n\\binom{n}{k}^2\\binom{2k}{k}(2k+1)^r(-1)^k.$$ In this paper we first prove the following congruences: $$\\sum_{k=0}^{n-1}S_k^{(2r)}\\equiv0\\pmod{n^2},$$ $$\\sum_{k=0}^{n-1}T_k^{(2r)}\\equiv0\\pmod{n^2}$$ and $$\\sum_{k=0}^{p-1}T_k^{(2)}\\equiv\\frac{p^2}{2}\\left(5-3\\big(\\frac{p}{5}\\big)\\right)\\pmod{p^3}.$$ We also show that there exist integers $a_{2r-1}$ and $b_r$, independent of n, such that $$a_{2r-1}\\sum_{k=0}^{n-1}S_k^{(2r-1)}\\equiv0\\pmod{n^2},$$ $$b_r\\sum_{k=0}^{n-1}kS_k^{(r)}\\equiv0\\pmod{n^2}.$$ This confirms several recent conjectures of Guo and Liu.",
  "revisions": [
    {
      "version": "v1",
      "updated": "2015-11-17T01:36:25.000Z"
    }
  ],
  "analyses": {
    "keywords": [
      "congruences",
      "conjectures",
      "positive integers"
    ],
    "note": {
      "typesetting": "TeX",
      "pages": 14,
      "language": "en",
      "license": "arXiv",
      "status": "editable"
    }
  }
}