{
  "id": "2210.15966",
  "version": "v1",
  "published": "2022-10-28T07:56:33.000Z",
  "updated": "2022-10-28T07:56:33.000Z",
  "title": "Some algebraic identity and its relations to Stirling numbers of second kind",
  "authors": [
    "Paweł Lorek"
  ],
  "categories": [
    "math.CO",
    "math.PR"
  ],
  "abstract": "In this short note we provide some algebraic identity with a proof exploiting its probabilistic interpretation. We show several consequences of the identity, in particular we obtain a new representation of a Stirling number of second kind, $$ S(n,d)={1\\over d!} \\sum_{1\\leq j_1<j_2<\\ldots<j_{d-1}< n} 1\\cdot2^{j_{d-1}-j_{d-2}}\\cdots d^{j_1}$$ for integers $n\\geq d$. Relating this to other known formula for $S(n,d)$ we also obtain $$ \\sum_{1\\leq j_1\\leq j_2\\leq \\cdots\\leq j_{n-d}\\leq d} j_1j_2\\ldots,j_{n-d} =d! \\sum_{1\\leq j_1<j_2<\\ldots<j_{d-1}< n} 1\\cdot2^{j_{d-1}-j_{d-2}}\\cdots d^{j_1}.$$ As a side effect, we have new proof of a known result stating that for any integer $d\\in\\mathbb{N}$ and any $x\\in\\mathbb{R}$ equality $$\\sum_{r=0}^d (-1)^r{d\\choose r}(x-r)^d=d!$$ holds. This is a special case of the presented identity.",
  "revisions": [
    {
      "version": "v1",
      "updated": "2022-10-28T07:56:33.000Z"
    }
  ],
  "analyses": {
    "subjects": [
      "11B73",
      "60C05",
      "11B83"
    ],
    "keywords": [
      "second kind",
      "algebraic identity",
      "stirling number",
      "short note",
      "special case"
    ],
    "note": {
      "typesetting": "TeX",
      "pages": 0,
      "language": "en",
      "license": "arXiv",
      "status": "editable"
    }
  }
}