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0 votes
1 answer
368 views

Identity involving Stirling number of the second kind

I'm looking for a citable reference for the following identity involving the Stirling numbers of the second kind $S(n, k)$ stated in Equation (27): For $n \geq 2$, $$ \sum_{m=1}^n S(n, m) (-1)^m (m-1)!...
Tim Seppelt's user avatar
7 votes
0 answers
444 views

Is there any literature on $\sum_{i=1}^{k} \left[ {k \atop i} \right] \zeta(i+1) $?

As per these questions, I'm trying to evaluate $$\sum_{n=2}^{\infty} \big{(} \zeta(n)^{2}-1 \big{)} = 1+ \sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m}. $$ Here, $H_{x}$ is a generalized Harmonic ...
Max Lonysa Muller's user avatar
1 vote
1 answer
211 views

Finite differences of Stirling numbers

Let s(n,k) and S(n,k) denote the Stirling numbers of the first (with signs) and second kinds, respectively. Next consider the sequence |s(n+2,n)| which begins: (2,11,35,85,175,...) . Using this to ...
user2052's user avatar
  • 1,411
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
1 vote
1 answer
299 views

Proof of Stirling number symmetric formulas [closed]

I'm looking for a reference to a proof of formulas 6.26 and 6.27 in Concrete Mathematics: $\def\sone#1#2{\left[#1\atop #2\right]} \def\stwo#1#2{\left\{#1\atop #2\right\}} $ $$ \stwo{n}{n-m} = \sum_k \...
user2263611's user avatar
9 votes
1 answer
568 views

Is this a new formula? $\Delta^d x^n/d! = \sum_k \left[ x \atop k\right]{ k+n \brace x + d}(-1)^{x+k}$

$$\frac{\Delta^d x^n}{d!} = \sum_k \left[ x \atop k\right]{ k+n \brace x + d}(-1)^{x+k}$$ Where $x$, $n$ and $d$ are non-negative integers, $\Delta^d$ is the $d$-th difference with respect to $x$, $\...
adam W's user avatar
  • 191