Questions tagged [stirling-numbers]
The stirling-numbers tag has no usage guidance.
14 questions with no upvoted or accepted answers
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 ...
- 4,829
5
votes
0
answers
775
views
A conjecture about the degrees of special polynomials
Define the congruence "modulo m" on exponential Taylor series as
$$
\sum_{n=0}^\infty \frac{a_n}{n!}x^n \equiv \sum_{n=0}^\infty \frac{b_n}{n!} x^n \mod m \iff \forall n: \frac{a_n-b_n}{m}\in \mathbb{...
- 237
4
votes
0
answers
312
views
Positivity of a finite sum involving Stirling numbers of the first kind
Past days I've been trying to prove that certain polynomials have positive coefficients. After a lot of thinking, I came up with a formula for each coefficient individually, and they are not that ugly....
- 1,889
4
votes
0
answers
158
views
Multiple integral evaluation involving Stirling numbers and Riemann zeta function
Hello Mathoverflow community, how are you doing? I just wanted to know if anything is known about the following integral:
$$K_n(m) = \overbrace{\int_0^1 \dots \int_0^1}^{n-\mathrm{times}} \left(-\...
- 1,549
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\...
- 2,232
3
votes
0
answers
89
views
Recursion for reversed rows of the A373183 using unsigned Stirling numbers of the first kind
Let $\left[{n \atop k}\right]$ be unsigned Stirling numbers of the first kind. Here
$$
\left[{n \atop k}\right] = (n-1)\left[{n-1 \atop k}\right] + \left[{n-1 \atop k-1}\right], \\
\left[{n \atop 0}\...
- 4,948
3
votes
0
answers
190
views
Stirling number, Delannoy number, and binomial coefficients in a sum
I want to compute/prove that the following sum is positive:
$$ \sum_{i = 0}^n \left[\frac{D(n - i, i)}{d} \sum_{j = m}^d s(d, j) \binom{j}{m} (d - i)^{j - m}\right] > 0
$$
where $s(d, j)$ is the ...
- 51
3
votes
0
answers
170
views
Stirling number bounds and polynomials and the Lambert $W$ function
Let $s(n,k)$ be the (signed) Stirling numbers of the first kind. The polynomials
$$L_n(x)=\sum_{j=1}^ns(n,n+1-j)\dfrac{x^j}{j!}$$
enter in the asymptotic expansion of the Lambert $W$ function, see for ...
- 13.1k
2
votes
0
answers
126
views
Generalized identity with Stirling numbers of the second kind and falling factorials
It is known that Striling numbers of the second kind satisfy the relation
$$
\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)\...
- 4,948
2
votes
0
answers
125
views
Inequality for 2-associated Stirling numbers of the second kind
Let $S_2(n,k)$ denote the 2-associated Stirling number of the second kind for $n$ objects and $k$ blocks, with $n$ being at least two. That is, we partition $n$ labeled objects into $k$ unlabeled ...
1
vote
0
answers
34
views
Closed form for $a(2^m(2k+1))$
Let $a(n)$ be A329369 (i.e., number of permutations of $\{1,2,\dotsc,m\}$ with excedance set constructed by taking $m-i$ ($0 < i < m$) if $b(i-1) = 1$ where $b(k)b(k-1)\cdots b(1)b(0)$ ($0 \...
- 4,948
1
vote
0
answers
86
views
Closed form for the family of polynomials
Let $s(n,k)$ be a (signed) Stirling number of the first kind.
Let $R(n,x)$ be the family of polynomials such that
$$
R(2n+1,x) = xR(n,x), \\
R(2n,x) = x(R(n,x+1) - R(n, x)), \\
R(0, x) = x
$$
Let $\...
- 4,948
1
vote
0
answers
59
views
Simple recursion for the A329369 using Stirling numbers of both kinds
Let $s(n,k)$ be a (signed) Stirling number of the first kind.
Let $n \brace k$ be a Stirling number of the second kind.
Let $a(n)$ be A329369 (i.e, number of permutations of ${1,2,...,m}$ with ...
- 4,948
1
vote
0
answers
63
views
On the equation involving Stirling numbers of the second kind ${n\brace a}{m\brace b}={k\brace c}$, and its solutions satisfying certain requirements
In this post we denote the Stirling numbers of the second kind as ${r\brace s}$ and we consider the proposal to ask if the equation of the title has infinitely many solutions $${n\brace a}{m\brace b}={...
