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2 votes
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Generalized identity with Stirling numbers of the second kind and falling factorials

It is known that Striling numbers of the second kind satisfy the relation $$ \sum\limits_{k=0}^{n}{n \brace k}(x)_k = x^n. $$ where $(x)_n$ is the falling factorials such that $$ (x)_n = x(x-1)(x-2)\...
Notamathematician's user avatar
3 votes
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Stirling number bounds and polynomials and the Lambert $W$ function

Let $s(n,k)$ be the (signed) Stirling numbers of the first kind. The polynomials $$L_n(x)=\sum_{j=1}^ns(n,n+1-j)\dfrac{x^j}{j!}$$ enter in the asymptotic expansion of the Lambert $W$ function, see for ...
Henri Cohen's user avatar
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