There are two kinds of Stirling numbers. Stirling numbers of the first kind $[{n \atop k}]$ count the number of ways to arrange $n$ objects into $k$ cycles. Stirling numbers of the second kind $\{ {n \atop k} \}$ count the number of ways to partition a set of $n$ objects into $k$ subsets.

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Stirling numbers: $S(n,k)=\sum_\limits{m=k}^n k^{n-m}S(m-1,k-1)$ [on hold]

How can I show $S(n,k)=\sum_\limits{m=k}^n k^{n-m}S(m-1,k-1)$ holds for the Stirling numbers, $n\geq m \geq k \geq 2$.
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28 views

Some proofs regarding Stirling numbers

I would like you to help me to prove two proofs correlated with Stirling numbers(the first one includes Stirling numbers of the second kind and the second one I guess Stirling numbers of the second ...
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70 views

Proof of an equation involving Stirling numbers of the second kind

I found this equation involving Stirling numbers of the second kind on Math World: $$\sum\limits_{m=1}^n (-1)^m(m-1)!\,S(n,m)=0 \ .$$ However, I do not know why this is true. I am looking for a proof ...
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97 views

An identity involving Bernoulli and Stirling numbers

I was playing with some combinatorial sums and made an observation that I didn't know how to prove: $$\forall n\in\mathbb N,\hspace{10px}\sum_{k=1}^n\frac{B_k\ ...
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42 views

Congruence for Stirling Number of first kind $s(n,k)$ when $n$ is prime

Let $s(n,k)$ be the Stirling numbers of first kind: $$\prod_{k=0}^{k=n-1}(x-k) =\sum_{k=0}^{k=n}s(n,k)x^k$$ $p$ is prime $\iff$ for all $k\in\{2,..,p-1\}$, $s(p,k)\equiv0\ mod\ p $ How ...
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97 views

Product of Stirling Numbers of the first kind

I have been messing around with coefficients of various polynomials and was wondering if there was a way to reduce the following stuff. Let polynomial, ...
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29 views

Final step of a random walk proof

I am working through the last bit of a random walk proof to show that a 3-d random walk is transient. The result I am looking for states that: $\frac {1}{2}^{2s} {{2s}\choose{s}} \sum_{j+k\leq{n}} ...
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Simplifying $\sum\limits_{k=0} {n \brack 2k}$ [closed]

Simplify $\sum\limits_{k=0} {n \brack 2k}$ where ${n \brack 2k}$ is Stirling's number of the first kind and $n \geq 0$
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31 views

What are the initial conditions for the Associated Stirling numbers of the second kind?

I'm trying to compute the Associated Stirling numbers of the second kind from the recurrence relation ...
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52 views

Sum involving Stirling numbers of the second kind

Is it possible to simplify this sum: $\sum\limits_{k=1}^{n} {n\brace k} (x)_k k =?$ Where, $ {n\brace k}$ are Stirling numbers of the second kind and $(x)_k$ if a falling factorial. Note: it is ...
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25 views

Approximation using Stirling

In the article "Series evaluation of Tweedie exponential dispersion model densities" by Peter Dunn and Gordon Smyth it is stated that $$-\log\Gamma(1+j)-\log\Gamma(-\alpha j)\approx ...
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115 views

Stirling numbers: Combinatorial proof of an identity

How to prove the following combinatorially ? \begin{equation} {n+1 \brace k+1}=\sum_{m=k}^{n}(k+1)^{n-m}{m \brace k}. \end{equation} My question is how are only ( n - m ) elements being considered ...
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60 views

How to prove this equality for Stirling numbers?

How can I prove that the following formula is true for Stirling numbers of first kind. $$\sum_{k=1}^n(-1)^k\left[\begin{matrix} n\\k\end{matrix}\right] =0$$ Actually I want to prove that number of ...
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1answer
29 views

Is this formula correct for the stirling numbers of the second kind?

If $S \left( n , k \right)$ is the number of partitions with $k$ blocks we can make from a set with $n$ elements, is the formula $$ S \left( n + 1, k \right) = \sum_{i = 0} ^{n} \binom{n}{i} S \left( ...
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1answer
41 views

Probability of picking each of m elements at least once after n trials.

Suppose I have 10^9 distinct elements, and an equal probability of picking each one in a given trial. How many trials must be conducted for the probability of having picked every element at least once ...
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377 views

Challenging identity regarding Bell polynomials

Note: [2015-03-08] A proof of the identity below was aimed to close the gap of a rather extensive elaboration of this answer of mine. The identity (1) below is part of a more complex one, which is ...
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84 views

Higher Order Terms in Stirling's Approximation

Some websites and books give stirling approximation as $$n! = \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \left( 1 + O \left(\frac{1}{n} \right)\right)$$ However when I check their derivations most ...
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1answer
58 views

Example of Stirling Numbers of the First Kind

I am trying to calculate the stirling numbers of the first kind. I am not very good in math and there is not a single example somewhere in the internet. So I would really appreciate it if you could ...
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1answer
68 views

Is there a more precise modified stirling's approximation formula for calculating n!?

I am trying to solve a problem of competitive programming Consider two integer sequences $f(n) = n!$ and $g(n) = a^n$, where $n$ is a positive integer. For any integer $a > 1$ the second ...
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41 views

Partitioning items into fixed size sets

I have the following problem: Given $n$ items, where each item has weight $w_{i}$, $i=1,2\ldots n$, what is the number of ways to partition these items into boxes of fixed size $C$, such that the sum ...
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69 views

Expansion for r-associated Stirling numbers of the second kind

I am looking for a paper or guidance for expanding the r-associated Stirling numbers of the second kind $S_r(n,k)$. $S_r(n,k)$ is the number of ways to partition a set of n objects into k subsets, ...
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63 views

Trying to prove a congruence for Stirling numbers of the second kind

I am struggling with a demonstration for this: When $n$ and $m$ are 2 natural integers such that $n-m$ is odd, then the following congruence holds for Stirling number of the second kind ${n \brace ...
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97 views

Finite sum involving Stirling numbers

I am trying to evaluate the following finite sum: $$ \sum_{h=0}^{m}\binom{m}{h}2^{m-h}S(h,k-r)S(m-h,r),\qquad 0\leq r\leq k\leq m, $$ where $S(n,k)$ is the Stirling number of the second kind. Can ...
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51 views

Find closed form formula for $c(n,n-4)$.

Find closed form formula for $c(n,n-4)$. Where $c(n,k)$ are signless stirling numbers of first kind. I need help.This is my last question all others problems of my exercise I have solved this is ...
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93 views

Find closed formula for $S(n,n-4)$.

Find closed formula for $S(n,n-4)$. These are stirling numbers of 2nd kind. My attempt: I uses the recurrence relation $$S(n,k)=S(n-1,k-1)+kS(n-1,k)$$ ...
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1answer
43 views

fixed length of permutaions cycles

How much permutations has only 10 cycles, but three of them has length 3 and seven of them has length 7?
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1answer
106 views

Prove the summation involving Stirling numbers of the first kind

I have been trying to prove or disprove this for 2 days now, but i don't even know where to begin. $$ 1 = \sum_{m=1}^{n} \sum_{k=1}^{n} \frac{x^{n-m}(-1)^{n-k-m} \left[\matrix ...
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3answers
157 views

My (divergent) summation of the zetas with sets of cofactors give systematically errors of simple integer differences. What am I missing?

This is a "fiddling" in a small project of mine with which I'm concerned from time to time for three years now. I try to focus on the core of the problem, please ask if more context is needed. ...
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83 views

Primality of Stirling numbers of second kind (again)

This question follows a previous one on the primality of Stirling numbers of the second kind ${n \brace k}$. Gerry indicated a paper on the topic. In this paper it is shown that for ${n \brace k}$ to ...
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38 views

Stirling numbers of the second kind — a series-expansion typo?

In H. S. Wilf's generatingfunctionology, (1.6.8) describes: $$ A_n(y) = \sum_k \begin{Bmatrix}n-1\\k-1\end{Bmatrix} y^k + \sum \begin{Bmatrix}n-1\\k\end{Bmatrix} y^k $$ $$ = yA_{n-1}(y) + \left( y ...
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68 views

Maximizing Stirling numbers of the second kind

In Stanley's Enumerative Combinatorics, there is a question on Chapter $1$ which goes as follows: Let $S(n,k)$ denote a Stirling number of the kind (ie, $S(n,k)$ is the number of ways to to ...
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57 views

Estimating the behavior for large $n$

I want to find how these coefficients increase/decrease as $n$ increases: $$ C_n = \frac{1}{n!} \left[(n+\alpha)^{n-\alpha-\frac{1}{2}}\right]$$ with $\alpha=\frac{1}{br-1}$ and $0\leq b,r \leq 1$. ...
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69 views

Combinatorial proof of an identity of Striling number of first kind

I can prove this identity using induction but i was looking for a combinatorial proof for this identity regarding stirling numbers of first kind. How should i proceed? Where, Thanks in advance.
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Primality of Stirling numbers of second kind

Apart from the Mersenne primes $M_p=2^p-1=\begin{Bmatrix}p+1\\2\end{Bmatrix}$, and the four primes $\begin{Bmatrix}n\\4\end{Bmatrix}$ where $n$ is given in http://oeis.org/A100958, are there other ...
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1answer
37 views

Total no. of partitions of $A$ using stirling no. of $2^{\text{nd}}$ kind

My notes describe Stirling numbers of $2^{\text{nd}}$ kind as : Given non-negative integers $r,n$ the stirling numbers of $2^{\text{nd}}$ kind denoted by are defined as: no. of ways of distributing ...
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Partial sums of Nicomachus' Triangle rows produce Stirling numbers of the 2nd kind?

I took partial sums of this triangle OEIS A036561 and found Stirling numbers of the 2nd kind. At OEIS A000392, at the mid-point of the comments section, is a conjecture. I think it's what I found. I ...
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1answer
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Does $\sum_{i = 1}^n S(k, i)i! = n^k$?

Consider the number of ways of ditributing $k$ distinct objects into $n$ distinct boxes, where $k \ge n$. On one hand, we can assign a box to each object. There are $n^k$ ways to do this. On the ...
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Stirling numbers and Power Group Enumeration

The following question is a reference request concerning a derivation of the EGF for the Stirling numbers of the second kind by Power Group Enumeration / Burnside's Lemma, which is $$\sum_{n\ge 0} ...
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1answer
126 views

Proving combinatorial identity with the product of Stirling numbers of the first and second kinds

$$ \sum_{k} \left[\begin{array}{c} n\\k \end{array}\right] \left\{\begin{array}{c} k\\m \end{array}\right\} = {n \choose m} \frac{\left( n-1\right)!}{\left(m-1 \right)!}, \quad \text{for } n,m > 0 ...
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1answer
78 views

Prove the identity involving summation and Stirling numbers of the second kind

Prove the identity $$(e^z-1)^m=m!\sum_{n}^{}{n \brace m}\frac{z^n}{n!}$$ $n\brace m$ stands for Stirling numbers of the second kind. I'm not really sure if $z$ is some special number or just an ...
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stirling numbers and harmonic number identities

Permit me a brief introduction before I state the question, three questions in fact. Inspired by this MSE link I computed the following harmonic sum identities: $$1/6\, \left( {H_{{n}}}^{(1)} \right) ...
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Generalization for Stirling numbers 2nd kind to negative column-indexes?

The exponential generating functions for the Stirling numbers 2nd kind are the n'th powers of $f(x)=\exp(x)-1$ (where this is understood as formal power series, Abramowitz&Stegun, 26.8.12). ...
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An asymptotic term for a finite sum involving Stirling numbers

The question is a by-product at the end of this post. The following asymptotic term will ensure the convergence of some series. $$ \frac{1}{n!} \sum_{k = 1 }^{n } \frac{{n \brack k}}{k+1} = ...
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1answer
71 views

Asymptotic approximation for the r-associated Stirling numbers of the second kind

It is well know that for fixed $k$ the asymptotic approximation for the Stirling numbers of the second kind is given by $\frac{k^n}{k!}$. Does such simple asymptotic expression also exist for the ...
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87 views

Stirling-like sum equal to zero when $k>n$

I need to prove that $$\sum_{r=0}^k\binom{k}{r}(-1)^r r^n=0$$ when $n<k$. I know that the formula above can be easily transformed into the Stirling number of the Second kind formula, which is ...
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1answer
78 views

Closed form for the Stirling numbers of the second kind.

I have realized through inspection that ${n \brace 2}=2^{n-1}-1$ and I have figured out with the help of Pedro Tamaroff that ${n \brace 3}=\frac{1}{6}(3^{n}-3\cdot2^n+3)$. For what other values of $k$ ...
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What is this notation, similar to the binomial coefficient?

I've come accross this notation: $$\left\{\begin{eqnarray}n\\m\end{eqnarray}\right\}$$ The only other info I have about this notation is that $\left\{\begin{eqnarray}4\\2\end{eqnarray}\right\}=7$ ...
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79 views

Stirling number of the second kind recurrence relations

I am interested to understand why the following recurrence relations of the Stirling number so the second kind hold using counting arguments ...
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1answer
100 views

Stirling Number of the Second Kind (intuition for formula)

The Stirling number of the second kind is the way of putting $n$ objects into $k$ nonempty boxes. I would like to understand the right hand side of this equation by a counting argument $$S(n,k) = ...
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How do I show the following identify where c(n , k) represent unsigned stirling numbers of the first kind?

I need to show the following identity but I don't know where to start. $ c(n+1, m+1) = \sum_{k=0}^{n} c(n, k) \binom{k}{m} $