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Questions tagged [stirling-numbers]

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3 votes
1 answer
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Sum with Stirling numbers of the second kind

Let $wt(n)$ be A000120, number of $1$'s in binary expansion of $n$ (or the binary weight of $n$) and $$n=2^{t_1}(1+2^{t_2+1}(1+\dots(1+2^{t_{wt(n)}+1}))\dots)$$ Then we have an integer sequence given ...
Notamathematician's user avatar
13 votes
4 answers
3k views

Simple/efficient representation of Stirling numbers of the first kind

Stirling numbers of the second kind can be expressed by means of a simple hypergeometric (considering $n$ fixed) sum $$S_2(n,k) = \frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}{k \choose j} j^n. \qquad (1)$$ ...
Fredrik Johansson'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
7 votes
2 answers
820 views

Determinant of matrix with Stirling numbers as elements

After noticing that the determinant of an $n \times n$ matrix $A_n$ with elements $a_{i,j}=i^j$, $1 \le i \le n$, $1 \le j \le n$, is the superfactorial (product of the first $n$ factorials), I wanted ...
Fabius Wiesner'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
4 votes
0 answers
578 views

A combinatorial bound involving Stirling numbers of the second type

My previous question was solved in a very elegant way, hopefully this (seemingly more complicated) case is also easy for experts. I need the inequality $\Big(\prod^r_{i=1}p_i\Big)\sum^n_{j=0}(-1)^j\...
Dmitry Kerner's user avatar
4 votes
1 answer
385 views

Relations involving Stirling numbers of second kind

While inverting a Laplace transform using Post's inversion formula I found the following expression: $$ \sum_{k=1}^n S^n_k \ x^k(\alpha)_k $$ where $S^n_k$ is a Stirling number of second kind and $(\...
guaraqe's user avatar
  • 157
4 votes
0 answers
158 views

Multiple integral evaluation involving Stirling numbers and Riemann zeta function

Hello Mathoverflow community, how are you doing? I just wanted to know if anything is known about the following integral: $$K_n(m) = \overbrace{\int_0^1 \dots \int_0^1}^{n-\mathrm{times}} \left(-\...
Brian's user avatar
  • 1,549
4 votes
0 answers
312 views

Positivity of a finite sum involving Stirling numbers of the first kind

Past days I've been trying to prove that certain polynomials have positive coefficients. After a lot of thinking, I came up with a formula for each coefficient individually, and they are not that ugly....
Luis Ferroni's user avatar
  • 1,889
2 votes
1 answer
345 views

Show that $\sum_{i=0}^{2k} [ {n\choose i+1} + (-1)^{i+1}{n+i+1\choose i+1} ] \sum_{j=0}^i {i\choose j}(-1)^j (i+1-j)^{2k} =0.$

Let $u(k,j) = 1$ if $j=0$, $0$ if $j > k$, or else it is $j*u(k-1,j-1) +(j+1)*u(k-1,j) $. Prove that $ \sum_{i=0}^{2k} {n \choose i+1} u(2k,i) +\sum_{i=0}^{2k} {-n-1 \choose i+1} u(2k,i)=0. $ ...
Benjamin L. Warren's user avatar