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2 votes
0 answers
126 views

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
5 votes
0 answers
775 views

A conjecture about the degrees of special polynomials

Define the congruence "modulo m" on exponential Taylor series as $$ \sum_{n=0}^\infty \frac{a_n}{n!}x^n \equiv \sum_{n=0}^\infty \frac{b_n}{n!} x^n \mod m \iff \forall n: \frac{a_n-b_n}{m}\in \mathbb{...
Danil Krotkov's user avatar