English

Some algebraic identity and its relations to Stirling numbers of the second kind

Combinatorics 2022-11-01 v2 Probability

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)=1d!1j1<j2<<jd1<n12jd1jd2dj1 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 ndn\geq d. Relating this to other known formula for S(n,d)S(n,d) we also obtain 1j1j2jnddj1j2,jnd=d!1j1<j2<<jd1<n12jd1jd2dj1. \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 dNd\in\mathbb{N} and any xRx\in\mathbb{R} equality r=0d(1)r(dr)(xr)d=d!\sum_{r=0}^d (-1)^r{d\choose r}(x-r)^d=d! holds. This is a special case of the presented identity.

Keywords

Cite

@article{arxiv.2210.15966,
  title  = {Some algebraic identity and its relations to Stirling numbers of the second kind},
  author = {Paweł Lorek},
  journal= {arXiv preprint arXiv:2210.15966},
  year   = {2022}
}

Comments

6 pages; minor typos corrected