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Questions tagged [eulerian-numbers]

For questions about the Eulerian numbers $A_{n,k}$, defined as the number of permutations in the symmetric group $S_n$ having $k$ descents. Not to be confused with Euler’s number $e$ or the Euler numbers $E_n$.

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3 votes
1 answer
369 views

Eulerian Polynomial Generating Function Proof

The generating function for the Eulerian polynomials is $$\frac{t-1}{t-e^{(t-1)x}} = \sum_{n=0}^\infty A_n(x) \frac{t^n}{n!}$$ where $A_n(x)$ is the $n^{th}$ Euler polynomial and $$A_n(x) = \sum_{k=...
3 votes
1 answer
79 views

The value of the second-order Eulerian polynomials at $x = \frac{-1}{2}$.

Recently, the second-order Eulerian polynomials $ \left\langle\!\left\langle x \right\rangle\!\right\rangle_n $ have been discussed on MSE [ a , b ]. $$ \left\langle\!\left\langle x \right\rangle\!\...
0 votes
0 answers
23 views

Request bibliographic reference(s) for finite alternating sums with Eulerian numbers

I would like to know a bibliographic reference for a math formula. I found this formula on Wikipedia, but no reference is given. It's the 2nd formula ($A(n,k)$ is an Eulerian number): $$\sum_{k=0}^{n-...
3 votes
2 answers
166 views

Sum of the series $\sum_{k=0}^\infty k^ne^{-k}$ for a positive integer $n$

How can we calculate the sum of the series $\sum_{k=0}^\infty k^ne^{-k}$ for a positive integer $n$? I tried: $$\sum_{k=0}^\infty k^ne^{-k} =(-1)^n \sum_{k=0}^\infty \frac{d^n}{dy^n}\bigg|_{y=1}e^{-ky}...
1 vote
1 answer
103 views

Identity connecting Stirling numbers of both kinds with second order Eulerian numbers

Setting. On page 270 of the great book Concrete Mathematics by Graham, Knuth and Patashnik, the second order Eulerian numbers $\langle\!\langle {n\atop k}\rangle\!\rangle$ with $n,k\in\mathbb{Z}$, $n \...
1 vote
2 answers
89 views

Does $\sum_{k=0}^{\infty} \sum_{n=1}^{\infty} x^nn^k =cx-x\ln(x)+x\ln(1-x) $ for $0 < x < 1$ for some real $c$? If so, what is $c$?

Let $g_k(x) =\sum_{n=1}^{\infty} x^nn^k $ and $G(x) =\sum_{k=0}^{\infty} g_k(x) =\sum_{k=0}^{\infty} \sum_{n=1}^{\infty} x^nn^k $. For $0 < x < 1$ is $G(x) =cx-x\ln(x)+x\ln(1-x) $ for some real $...
1 vote
3 answers
129 views

Power series of $x/(1-ae^{-x})$.

I am looking for a power series expansion for the function $x(1-ae^{-x})^{-1}$ (perhaps for $0<a<1$). Using the Bernoulli numbers, I can write \begin{align*} \frac{x}{1-ae^{-x}} &= \frac{x}{...
4 votes
3 answers
364 views

Recurrence formula for the Eulerian and derangement polynomial

For the Eulerian polynomial $$A_n(x)=\sum_{\pi\in S_n}x^{\mathrm{des}(\pi)}$$ it is well known, that we have the nice recurrence formula $$A_n(x)=\sum_{k=0}^{n-1}\binom{n}{k}A_k(x)(x-1)^{n-1-k}.$$ I ...
6 votes
1 answer
214 views

The relation of the Bernoulli numbers to the Catalan numbers

The Bernoulli numbers $B_n$ are the backbone of calculus, and according to B. Mazur, they "act as a unifying force, holding together seemingly disparate fields of mathematics." The Catalan ...
2 votes
0 answers
97 views

Understanding the recurrence relation $f(n,k) = kf(n-1, k) + (n-k+1)f(n-1,k-1)$ for the Euler numbers $f(n,k)$

The Euler numbers $f(n,k)$ are given in Generatingfunctionology as the number of permutations of $[n]$ with exactly $k$ increasing runs (exercise 1.18.c). Its recurrence relation is $$f(n,k) = kf(n-1, ...
6 votes
1 answer
181 views

Series power function over exponential function

A typical exercise from calculus is to show that any exponential function eventually grows faster than any power function, i.e. $$ \lim_{k \to \infty} \frac{k^a}{b^k} = 0 \qquad \text{ for } a,b>1.$...
0 votes
2 answers
148 views

Proof that $e^{i\theta}/e^ {i\phi} = e^{i(\theta - \phi)}$ [closed]

Could anyone please help me with proving that \begin{equation} \frac{e^{i\theta}}{e^{i\phi}}= e^{i(\theta-\phi)} \end{equation} Is $e^{i\theta} = \cos(\theta)+i\sin(\theta)$ useful in this proof? ...
9 votes
2 answers
1k 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 ...
3 votes
3 answers
273 views

Simplify binomial sum

It is well-known that for any integer $j$ $$ \sum _{i=0}^k (-1)^{i} \binom{k}{i} (j-i)^k = k! $$ But I wonder if there's a closed form for the following sum $$ \sum _{i=j}^k (-1)^{i} \binom{k}{i} (j-i)...
0 votes
0 answers
78 views

Closed form of Eulerian numbers Proof

I was reading the proof of the closed-form formula of Eulerian numbers ($A(n,k)$) from Bona's book. I had a doubt in his classification of "removable fences" and how they eliminate the ...
1 vote
1 answer
101 views

Connecting Euler Numbers of the Second Kind and Unsigned Stirling Numbers of the First Kind. [closed]

The notation $ \left\langle\left\langle n\atop m\right\rangle\right\rangle$ denotes the Eulerian numbers of the second kind or order. Similarly, ${n \brack m}$ denotes the unsigned Stirling numbers of ...
-1 votes
1 answer
303 views

Worpitzky's identity eulerian number - formula demonstration [closed]

Can someone tell me please how am I supposed to demonstrate the Worpitzky's identity : $$ x^{n}=\sum_{k}\left\langle\begin{array}{l} n \\ k \end{array}\right\rangle\left(\begin{array}{c} x+k \\ n \end{...
8 votes
1 answer
299 views

Bounds on the difference between the polylogarithm with negative base and the gamma function

Trying to understand intuitively the Gamma function I started to think of it as a way to measure how much each factorial power "helps" $x^n$ in the infinite sum of $e^x$, thus trying to ...
5 votes
0 answers
71 views

Connection between Eulerian numbers and number of elements in set of uniform variables greater than the mean?

I was recently investigating the following question: Given $n$ independent $\text{Unif(0, 1)}$ variables $U_1,\ldots,U_n$, let $m$ be the number of elements in $[U_1,\ldots,U_n]$ that are greater than ...
6 votes
2 answers
141 views

An identity related to the second-order Eulerian numbers.

Recently, some of the remarkable properties of second-order Eulerian numbers $ \left\langle\!\!\left\langle n\atop k\right\rangle\!\!\right\rangle$ A340556 have been proved on MSE [ a , b , c ]. ...
3 votes
0 answers
61 views

The second-order Eulerian numbers meet the Clausen numbers

The (generalized) Clausen numbers A160014 are defined as $$\operatorname{C}_{n, k} = \prod_{ p\, -\, k\, |\, n} p \quad (p \in \mathbb{P})$$ where $\mathbb{P}$ denotes the primes. The classical ...
4 votes
1 answer
144 views

A recurrence of the second-order Eulerian polynomials

Recently, some of the remarkable properties of second-order Eulerian numbers $ \left\langle\!\!\left\langle n\atop k\right\rangle\!\!\right\rangle$ A340556 have been proved on MSE [ a , b , c ] ...
4 votes
2 answers
188 views

Difference of the Stirling cycle numbers and the Stirling set numbers

Denote by $\left\langle\!\! \left\langle k\atop j\right\rangle\!\! \right\rangle$ the second-order Eulerian numbers A340556. Define $$ \left| n\atop k\right| = \sum_{j=0}^k \left( \binom{n + j - 1}{...
6 votes
1 answer
312 views

Second-order Eulerian numbers, Lambert's W function, and Schröder's fourth problem

Recently, some of the remarkable properties of second-order Eulerian numbers $ \left\langle\!\!\left\langle n\atop k\right\rangle\!\!\right\rangle$ A340556 have been proved on MSE [ a , b , c ]. ...
7 votes
0 answers
255 views

Proof for a summation-procedure using the matrix of Eulerian numbers?

I've discussed a procedure for divergent summation using the matrix of Eulerian numbers occasionally in the last years (initially here, and here in MSE and MO but not in that generality and thus(?) ...
3 votes
1 answer
187 views

An associated Stirling number identity related to the second-order Eulerian numbers.

A similar Stirling number identity representing the second-order Eulerian numbers can be found at this question. We denote the associated Stirling cycle numbers as $\left[\!\left[ n\atop k\right] \! \...
5 votes
1 answer
303 views

A Stirling number identity representing the second-order Eulerian numbers.

Graham, Knuth, and Patashnik give in CMath a thorough introduction to the Stirling numbers. On table 250 and table 251, they compile two pages of Stirling number identities. Of course, there are many ...
2 votes
1 answer
186 views

A combinatoric identity involving Eulerian and Stirling numbers

Let $$ A_{n}(x)=\sum_{k=0}^{n}\left\langle\begin{array}{l} n \\ k \end{array}\right\rangle x^{n-k} $$ Eulerian Polynomials and $$ \left\langle\begin{array}{l} n \\ k \end{array}\right\rangle $$ are ...
2 votes
0 answers
51 views

Does the sum $\sum_{k\ge 0}A(n,k)k^t$ have a closed form?

My friend and I are trying to find a closed form for the sum $\sum_{k\ge 0}A(n,k)k^t,$ in which the notation $A(n,k)$ is for the Eulerian number. (An alternating notation is $\left<n\atop k\right&...
4 votes
2 answers
850 views

A closed form of Eulerian numbers

The following identity involving Eulerian numbers is well-known: \begin{equation} A(n,m)=\sum_{k=0}^{m}(-1)^k \binom{n+1}{k} (m+1-k)^n. \end{equation} where $A(n,m)$ is the number of permutations $(\...
4 votes
0 answers
104 views

About $t$-analogue of the Eulerian polynomials.

A certain way to define the $t$-analogue of the Eulerian polynomials $C_n(x)$ is by $$ C_n(x,t)=\sum_{\pi\in S_n}x^{\text{des}(\pi)+1}t^{\text{maj}(\pi)} $$ where $des(\pi)$ is the descents in $\pi$, ...
5 votes
2 answers
278 views

Why does $\frac{1}{2}A_n(2)$ count the number of ordered set partitions?

Suppose $A_n(x)$ denotes the Eulerian polynomial. Is there a combinatorial proof that $\frac{1}{2}A_n(2)$ counts the number of ordered set partitions? By this I mean a set partition of a set of $n$ ...
1 vote
1 answer
130 views

Probability generating function.

I have this collection of generating functions in $x$ parametrized by $n$: $$f_n(x) = \frac{1}{n!}(1-x)^{n+1}\sum_{k \geq 1} k^n x^k,$$ so that $n$ is a constant. For instance $$f_6(x) = (1/720)x+(...
1 vote
1 answer
161 views

Recurrence $C_n(t)=t(1-t)C'_{n-1}(t)+ntC_{n-1}(t)$ for Eulerian Polynomials?

I was reading about Eulerian polynomials on OEIS, and there is a recurrence given for them, namely: $$ C_0(t)=1 $$ and $$ C_n(t)=t(1-t)C'_{n-1}(t)+ntC_{n-1}(t)\qquad (n\geq 1). $$ How can ...
1 vote
0 answers
38 views

Sign alternating for $n$-th Eulerian polynomial at $x=-2$

What is the law of sign alternating for the $n$-th Eulerian polynomial evaluated at $x=-2$ (A087674)? I have some ideas, but need more (than 100 from related A212846) terms. How and where can I ...
0 votes
0 answers
62 views

Identity on the q-analog of the Eulerian polynomial

We define the q-analog of the Eulerian polynomial $A_n(x)$ as $A_n(x,q) = \sum_{\sigma \in S_n} x^{d(\sigma)+1}q^{maj(\sigma)}$, where $S_n$ is the set of permutations of $n$ elements, $d(\sigma)$ is ...
7 votes
1 answer
790 views

A nicer recurrence for the Eulerian polynomials.

I was perusing the subject of Eulerian polynomials. I'm assuming the definition that the Eulerian polynomial is defined by $C_n(t)=\sum_{\pi\in S_n}t^{1+d(\pi)}$, where $d(\pi)$ is the number of ...
1 vote
0 answers
117 views

Formula for $q$-Analogue of Eulerian Polynomial

Define $\text{d}(\sigma)$ and $\text{maj}(\sigma)$ to be the number of descents and the major index of the permutation $\sigma$, respectively. Define a $q$-Analogue of the Eulerian Polynomials by $A_n(...
2 votes
0 answers
194 views

Simon Newcomb's problem

I am looking for an answer to the following problem. Let $S$ be the multiset $\{1^{d_1},2^{d_2},\dots,m^{d_m}\}$ $A_{S,k}$ is the number of permutations of $S$ with $k-1$ descents and no descent at ...
3 votes
1 answer
60 views

Is my use of derivatives correct in this infinite sum?

I was inspired by a question that seeked to find a generating function for the sum of powers. We set $s(n,p)=\sum_{k=0}^n k^p$, now we are looking for $G(x,p)= \sum_{n=0}^{\infty}s(n,p)x^n$. ...
1 vote
1 answer
137 views

Number of Permutation with Pro = k equals the Eulerian Number A(n,k+1)

The problem phrases as follows: Let $\sigma = (\sigma_1, . . . , \sigma_n)$ be a permutation. We say that element $i$ is progressive if $\sigma_i > i$. We write $pro(\sigma)$ for the number of ...
1 vote
0 answers
753 views

Prove Eulerian Number using Combinatorics.

For any $n ≥ 1$ and $1 ≤ k ≤ n$, define the “Eulerian number” $e(n, k)$ to be the number of permutations of $\{1, 2, . . . , n\}$ with exactly $k −1$ descents. So $e(n, 1) = e(n, n) = 1$, and $e(n, k) ...
1 vote
1 answer
240 views

Eulerian Number Asymptotics

The Eulerian Number $A(n,k)$ is the number of permutations of $1$ to $n$ with exactly $k$ rises (or ascents, i.e., positions $i$ such that $x_i<x_{i+1}$). The article The Boundary of the Eulerian ...
3 votes
1 answer
1k views

Half of the binomial theorem

The binomial theorem states that the generating function $\sum_{k=0}^n {n \choose k} x^k$ is equal to $(1+x)^n$ for any $n$. For a given $n$, let $$B(x)=\sum_{k=0}^n {2n+1\choose k} x^k.$$ That is, ...
4 votes
1 answer
181 views

References/Proof of the conjectured identity for the Stirling permutation number $\left\{{n\atop n-k}\right\}$

While working with a combinatorics problem, I conjectured that $$ \left\{{n \atop n-k }\right\}=\sum_{p=0}^{k-1}\bigg\langle\!\!\bigg\langle{k\atop k-1-p}\bigg\rangle\!\!\bigg\rangle \binom{n+p}{2k}, ...
3 votes
2 answers
221 views

How to prove Eulerian Identity?

This is a question in the book Concrete Mathematics of Graham and Knuth about Eulerian numbers, Stirling Numbers of the second kind and binomial coefficients. I have been thinking about it for a lot ...
10 votes
1 answer
390 views

Positivity of a certain sum of Stirling numbers

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....
2 votes
1 answer
87 views

Proof that the Eulerian Numbers $A(n,m)$ are asymptotic to $(m+1)^n$ when $m$ is fixed and $n\rightarrow\infty$

This is stated in many places but seldom proved. The goal is to prove that, for fixed $m\geq 0$, $$\lim_{n\rightarrow\infty}A(n,m)\sim (m+1)^n$$ where $A(n,m)$ are the Eulerian Numbers, namely, the ...
3 votes
0 answers
166 views

Permutation statistics in multiple rows

Usually we study the statistics of a permutation written in one row. Is there any result for the statistics of a permutation written in multiple rows? Let me give an example in order to be more clear: ...
8 votes
0 answers
231 views

odd property of Eulerian numbers

One of the curious features of Pascal's triangle is that each row contains a two power number of odd entries. In fact, the precise number is $2^{b(n)}$ where $b(n)$ denotes the sum of the bits of $n$...