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arXiv:2210.15966 [math.CO]AbstractReferencesReviewsResources

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

Paweł Lorek

Published 2022-10-28Version 1

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.

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