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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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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 ...
Peter Luschny's user avatar
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] \! \...
Peter Luschny's user avatar
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 ...
omegadot's user avatar
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6 votes
1 answer
310 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 ]. ...
Peter Luschny's user avatar
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}{...
Peter Luschny's user avatar
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)...
LeafGlowPath's user avatar
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8 votes
1 answer
298 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 ...
Alejandro Quinche's user avatar
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(?) ...
Gottfried Helms's user avatar
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 ...
bronko's user avatar
  • 265
10 votes
1 answer
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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....
Luis Ferroni's user avatar
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 ...
Clara's user avatar
  • 1,536
5 votes
0 answers
223 views

Interpretation of Eulerian numbers using the principle of inclusion-exclusion

Eulerian numbers, denoted $e_{n,k}$, are defined as the number of permutations of $[n]$={1,2,...,n} such that there are k "ascents". (For example, the permutation 23541 of [5] would have 3 ascents, ...
settheorynoob's user avatar
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}, ...
Sangchul Lee's user avatar
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 ] ...
Peter Luschny's user avatar
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\!\...
Peter Luschny's user avatar
3 votes
1 answer
154 views

Does $x\left(\frac{d}{dx}\left(\cdots x \left(\frac{d}{dx} \left( \frac{x}{1-x}\right)\right)\cdots\right)\right)$ have a closed form expression?

Does $\displaystyle \underbrace{x\left(\dfrac{d}{dx}\left(\cdots x \left(\dfrac{d}{dx} \left( \dfrac{x}{1-x}\right)\right)\cdots\right)\right)}_{\text{$x \frac{d}{dx}$ $m$ times}}$ have a closed form?...
Puzzled417's user avatar
  • 6,956
2 votes
2 answers
375 views

Identities for sums of Eulerian numbers

Let $[n] := \{1,2,\dots,n\}$. An Eulerian number, $A(n,k)$, denotes the number of permutations $\sigma \in S_n$ of $[n]$ such that exactly $k$ numbers have the property that $\sigma(i) < \sigma(i+...
Celal Bey's user avatar
  • 516
1 vote
3 answers
128 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}{...
Kenneth Ng's user avatar
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 \...
azimut's user avatar
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1 vote
0 answers
58 views

Somehow "mirroring" the Taylor-expansion of some $g(x)$

In proceeding with an older discussion of a summation-procedure for divergent series using the matrix of Eulerian numbers I came to a rather general formulation but cannot/do-not-know-how-to make sure,...
Gottfried Helms's user avatar
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 $...
marty cohen's user avatar
1 vote
0 answers
152 views

Is that series-transformation known in the context of divergent summation

Background: In the context of divergent summation I'm analyzing the matrix of eulerian numbers for a regular matrix-summation method. Beginning indexes at zero (r for "row", c for "column") the ...
Gottfried Helms's user avatar