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### Why are these two gamma functions equal? [on hold]

$\gamma_{1}(s)=\pi^{\frac{1}{2}-s}\frac{\Gamma(\frac{s}{2})}{\Gamma(\frac{1}{2}(1-s))}$
$\gamma_{2}(s)=(\frac{b}{2\pi})^{\frac{1}{2}-a} e^{-2i\theta(b)}$
,where $s=a+bi$
I calculated ...

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### Closed form for Stirling numbers of second kind [closed]

so yeah, a friend and I have been trying to find a closed form for the Stirling numbers of second kind with little success.
https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind
The ...

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### 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 ...

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### Generating function for reciprocals of Stirling numbers?

Is there an (ordinary or exponential) generating function for the reciprocals $\frac{1}{s(n,k)}$ of the Stirling numbers of the first kind?
Also, is there some general way to find generating ...

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### 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 ...

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### 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 ...

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### 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 ...

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256 views

### Simple approximation to a sum involving Stirling numbers?

I have also posted this question at http://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:
...

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### How this expression leads to the given sequence

Here given is a sequence from OEIS.
The sequence is triangle of coefficients from fractional iteration of $e^x - 1$. Few terms are:
1, 1, 3, 1, 13, 18, 1, 50, 205, 180, 1, 201, 1865, 4245, 2700, 1, ...

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### 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 ...

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### 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 ...

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434 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 ...

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### 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:
...

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### Acyclic orientations of complete graphs in terms of Stirling numbers?

It is well-known that the number of acyclic orientations of $K_n$ is $n!$. Does anybody know of a combinatorial argument for this fact which uses the identity:
$$n!=\sum_{k=1}^ns(n,k),$$
where the ...

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### Stirling number identity via homology?

This is a question about the well-known formula involving both types of Stirling numbers:
$\sum_{k=1}^{\infty}(-1)^{k}S(n,k)c(k,m)=0$,
where $S(n,k)$ is the number of partitions of an $n$-element set ...

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### 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
...

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### 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} ...