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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What do the Stirling numbers of the first kind have to do with polylogarithms?

On a whim, I had decided to look into ways of evaluating series of the form $$\sum_{n\ge1}\frac{1}{n^k2^n}$$ which I learned has a more general form in terms of polylogarithms: ...
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Why is $\binom{2n}{n} \asymp \Theta \big(\frac{2^{2n}}{\sqrt{n}}\big)$?

I saw this statement : $$\binom{2n}{n} \asymp \Theta \bigg(\frac{2^{2n}}{\sqrt{n}}\bigg) \asymp \Theta\bigg(\frac{4^n}{\sqrt{n}}\bigg)$$ How did we go from the first statement to the second? I tried ...
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Stirling numbers of the second kind, general formula

I found out about the Stirling numbers (first and second) when I studied a way to smooth the factorial function. This is the way I want to define them. Approach the factorial $(1+n)!$ as a product and ...
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31 views

Simplify the formula for the number of distributions leaving none of the $n$ cells empty

I'd like to help with the following problem: $$ \binom{x}{r-1} + \binom{x}{r} = \binom{x+1}{r} \tag{8.6}\label{8.6} $$ 7. Let $A(r, n)$ be the number of distributions leaving none of the n cells ...
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Finding closed form for this summation

recently i have beeen asking alot of questions about summations, But this one is actually quite interesting: $$ \sum_{j=k}^n j! 2^{k-2j} \left({2j-k-1 \choose j-1} - {2j-k-1 \choose j}\right){n \brack ...
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Stirling number Combinatorics. Summation .

$$ \sum_{k=0}^n \left\{ {n\atop k} \right\} *(x)_k = x^n $$ is well known . What if the k-th term of LHS summation is divided by $q^k$ where $q$ is some positive constant, What about $$ ...
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125 views

Stirling transform of $(k-1)!$

While reading about combinatorial mathematics, I found this article about the Stirling transform which caught my attention. So, if I wanted to find the Stirling transform of, for instance, $(k-1)!$, ...
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34 views

How to show that the displaying numbers of a onto function is k!S(n,k)?

Let it be $A$,$B$ sets that $|A|$=$n$, $|B|$=$k$ and $|A|>|B|$. How to show that the displaying numbers of an onto function $f$:$A$ $\rightarrow$ $B$ is: $\begin{Bmatrix} n \\ k\end{Bmatrix}$$k!$ ...
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29 views

Counting functions and stirling numbers

Let S= { f | f: A $\rightarrow$ B, |Image(f)|=k}. |A|=m, |B|=n. where k $ \le n, k \le m $ |S|=$ {n \choose k} $ S(m,k) k!. where S(m,k) are the striling numbers of the second kind. What I can't ...
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59 views

What is the combinatorial proof for the formula of S(n,k) - Stirling numbers of the second kind?

What is the combinatorial proof for the formula of Stirling numbers of the second kind ? i.e. S(n,k) where n is the number of objects and k is the number of parts $${n\brace ...
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32 views

Falling power of a sum in terms of falling powers of the terms

I am trying to come up with an expression for $(x+y)^{\underline{n}}$ in terms of $x^{\underline{r}}$ and $y^{\underline{r}}$. I tried for $n=2$ and $n=3$ and it looks like binomial expansion holds, ...
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59 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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4answers
86 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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130 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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57 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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100 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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34 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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1answer
32 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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63 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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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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122 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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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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35 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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51 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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398 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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120 views

Polylogarithms of negative integer order

The polylogarithms of order $s$ are defined by $$\mathrm{Li}_s (z) = \sum_{k \geqslant 1} \frac{z^k}{k^s}, \quad |z| < 1.$$ From the above definition, derivatives for the polylogarithms ...
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99 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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82 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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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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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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78 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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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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103 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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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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45 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
115 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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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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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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1answer
42 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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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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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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1answer
71 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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45 views

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
39 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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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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127 views

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
131 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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84 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 ...