{
  "id": "1607.06366",
  "version": "v1",
  "published": "2016-07-21T15:37:27.000Z",
  "updated": "2016-07-21T15:37:27.000Z",
  "title": "On the equation $ n_1 ... n_{k+1} =n_{k+2}... n_{2(k+1)} $",
  "authors": [
    "Sanying Shi",
    "Michel Weber"
  ],
  "comment": "15 pages",
  "categories": [
    "math.NT"
  ],
  "abstract": "Let $k\\ge 1$ and let $N_{k}(B)$ denote the number of solutions of the equation $n_1\\ldots n_{k+1} =n_{k+2}\\ldots n_{2(k+1)}$ with unknowns verifying $1\\le n_i\\le B$,$1\\le i\\le 2(k+1)$. When $k=1$, very precise estimates are known. When $k=2$, we show that $B^3(\\log^3 B)(\\log\\log B)^{-1}\\ll N_2(B )\\ll B^3 \\log^4 B$. For $k>2$ we also show that $B^{k+1}(\\log B)^{[\\log_2k]}\\ll_k N_{k}(B) \\ll_k B^{k+1}\\big(\\log B\\big)^{k^2+2k-2}$. We further study the number of solutions $N(B,F)$ of the equation $n_1n_2=n_3n_4$, where $1\\le n_1\\le B$, $1\\le n_3\\le B$, $n_2, n_4\\in F$ and $F\\subset [1,B]$ is a factor closed set. Let $F= \\big\\{m=p_1^{\\epsilon_1}\\ldots p_k^{\\epsilon_k}, \\ \\epsilon_j\\in \\{0,1\\}, 1\\le j\\le k\\big\\}$, where $p_1<\\ldots<p_k$ are prime numbers, and let $B\\ge p_1p_2\\cdots p_k$. We prove that $ 2^kB\\big(1+2\\sum_{n\\in F\\atop n\\neq 1} \\frac{1}{2^{ \\omega(n)} n } \\big)\\ \\le \\ N(B,F)\\le B\\big(\\frac{5}{2}\\big)^{k }\\big( 1+ C \\sum_{n\\in F\\atop n\\neq 1} \\frac{1}{5^{ \\omega(n)} n } \\big)+ 4^k$, where $\\omega(n)$ is the prime divisor function and $C$ is a numerical constant.",
  "revisions": [
    {
      "version": "v1",
      "updated": "2016-07-21T15:37:27.000Z"
    }
  ],
  "analyses": {
    "subjects": [
      "11D57",
      "11A05",
      "11A25"
    ],
    "keywords": [
      "prime divisor function",
      "precise estimates",
      "factor closed set",
      "prime numbers",
      "numerical constant"
    ],
    "note": {
      "typesetting": "TeX",
      "pages": 15,
      "language": "en",
      "license": "arXiv",
      "status": "editable"
    }
  }
}