Questions tagged [stirling-numbers]
The stirling-numbers tag has no usage guidance.
63 questions
5
votes
0
answers
775
views
A conjecture about the degrees of special polynomials
Define the congruence "modulo m" on exponential Taylor series as
$$
\sum_{n=0}^\infty \frac{a_n}{n!}x^n \equiv \sum_{n=0}^\infty \frac{b_n}{n!} x^n \mod m \iff \forall n: \frac{a_n-b_n}{m}\in \mathbb{...
- 237
3
votes
1
answer
585
views
Trying to prove a congruence for Stirling numbers of the second kind
This a repost of a question I asked at Stack Exchange, but I got no answer so far, so I am trying here, even though it may not suit the "research level" requirement.
Proposition: When $n$ and $m$ are ...
- 505
9
votes
1
answer
825
views
An infinite set of identities using Stirling numbers 1st kind - are they all zero?
I have the following set of series involving the Stirling numbers 1'st kind and binomials, which can be understood as a set of dot-products of row- and column-vectors of two infinite matrices (where R ...
- 5,342
4
votes
0
answers
158
views
Multiple integral evaluation involving Stirling numbers and Riemann zeta function
Hello Mathoverflow community, how are you doing? I just wanted to know if anything is known about the following integral:
$$K_n(m) = \overbrace{\int_0^1 \dots \int_0^1}^{n-\mathrm{times}} \left(-\...
- 1,549
2
votes
1
answer
338
views
Simple approximation to a sum involving Stirling numbers?
I have also posted this question at https://math.stackexchange.com/questions/486917/simple-approximation-to-a-sum-involving-stirling-numbers. I have an exact answer to a problem, which is the function:...
- 23
5
votes
1
answer
585
views
Alternating sums of alternate Stirling numbers
Does anybody know of any identities or combinatorial interpretations for alternating sums of alternate Stirling numbers?
I am particularly interested in expressions of the form:
$$\pm\sum_{k}(-1)^k|...
- 1,327
1
vote
1
answer
287
views
Sum of Stirling numbers with exponents
I have a trouble with the following sum
$\sum_{i=0}^n\binom{n}{i}S(i,m)3^i$, where $S(i,m)$ is the Stirling number of the second kind (the number of all partitions of $i$ elements into $m$ nonempty ...
- 153
9
votes
1
answer
568
views
Is this a new formula? $\Delta^d x^n/d! = \sum_k \left[ x \atop k\right]{ k+n \brace x + d}(-1)^{x+k}$
$$\frac{\Delta^d x^n}{d!} = \sum_k \left[ x \atop k\right]{ k+n \brace x + d}(-1)^{x+k}$$
Where $x$, $n$ and $d$ are non-negative integers, $\Delta^d$ is the $d$-th difference with respect to $x$, $\...
- 191
4
votes
1
answer
886
views
A bound involving Stirling numbers of the second kind and the asymptotics
Let $S_{n,r}$ denote the Stirling number of the second kind. Define $A_{n,r}:=\frac{\binom{n+r-1}{n}(n+r)!}{S_{n+r,r}r!}$. I want to prove:
$A_{n,1}\ge A_{n,2}\ge..\ge A_{n,r}\ge \lim_{r\to\infty} ...
- 2,232
0
votes
2
answers
187
views
Combinatorial Interpretation of Generalized Stirling numbers
I know the combinatorial interpretation of first, and second order Stirling numbers (#of k cycles of n items, and #of partitions n items into k subsets). Is there an interpretation for the generalized ...
- 113
4
votes
1
answer
385
views
Relations involving Stirling numbers of second kind
While inverting a Laplace transform using Post's inversion formula I found the following expression:
$$
\sum_{k=1}^n S^n_k \ x^k(\alpha)_k
$$
where $S^n_k$ is a Stirling number of second kind and $(\...
- 157
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
0
votes
1
answer
194
views
Asymptotic formula for an expression in terms of the second kind of stirling numbers
We have proved that
the limit of $\sum_{k=0}^n r^2k^m / (1+r)^{k+1}$ when n approaches infinity is $\sum_{k=1}^m S(m,k)k!/r^{k-1}$
where S(m,k) is the second kind of stirling number.
Is there a ...
- 63