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

TL;DR

This paper presents a new algebraic identity with a probabilistic proof, leading to novel representations of Stirling numbers of the second kind and related summations, along with a new proof of a known combinatorial result.

Contribution

It introduces a new algebraic identity and derives new formulas for Stirling numbers of the second kind, connecting algebraic, combinatorial, and probabilistic perspectives.

Findings

New representation of Stirling numbers of the second kind.

Derived summation formulas involving Stirling numbers.

Provided a new proof of a known combinatorial identity.

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.