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5 votes
1 answer
230 views

Reference request: Gessel interview's generating function identities

In this interview, Ira Gessel mentions the following results: Result 1: Let $B_n$ denote the $n^{\text{th}}$ Bernoulli number. Define the series $$B(x) = \sum_{n=2}^{\infty} \frac{B_nx^{n-1}}{n(n-1)}.$...
Naysh's user avatar
  • 557
4 votes
0 answers
208 views

Extract this constant term

Given a Laurent polynomial $F$ in the variables $\mathbf{t}=(t_1,\dots,t_n)$, let $CT_{\vec{\mathbf{t}}}\,F$ denote its constant term. For example, $CT_{t_1,t_2}((8t_1-\frac1{3t_1t_2})(5t_1t_2+t_2^2+\...
T. Amdeberhan's user avatar
3 votes
2 answers
463 views

Ask for a reference or a proof of a combinatorial identity $\sum_{k=0}^n\binom{2n+1}{2k}\binom {k}{m} =2^{2(n-m)}\frac{2n+1}{2(n-m)+1}\binom{2n-m}{m}$

Could you please recommend a reference to or supply a proof of the following identity \eqref{combin-ID-Maclaurin}, or \eqref{first-equiv-form}, or \eqref{combin-ID-Mac-Equiv}, or \eqref{combin-ID-Mac-...
qifeng618's user avatar
  • 1,091
4 votes
1 answer
147 views

$0,1$-matrices with $1$ in every row/column vs. all $0,1$-matrices

Chapter 2, Exercise 25 of R. Stanley's "Enumerative Combinatorics" Vol. 1 asserts that $$ \sum_{m,n \geq 0} \left(\sum_{t \geq 0} f_i(m,n)t^i\right)\frac{x^m}{m!}\frac{y^n}{n!} = e^{-x-y}\...
Sam Hopkins's user avatar
  • 24.2k
6 votes
1 answer
166 views

An identity for rational functions leading to equations for multiple polylogarithms

The following identity is not hard to prove: $$ \sum_{1\leq i_1<i_2<\ldots <i_{2n}\leq N} (-1)^{i_1+\ldots+i_{2n}}\frac{(1-x_{i_1})(1-x_{i_3})\ldots(1-x_{i_{2n-1}})}{(1-x_{i_2})(1-x_{i_4}) \...
Daniil Rudenko's user avatar
11 votes
2 answers
1k views

Proofs of some combinatorial identities

Just wondering if anyone knows any references in the literature to bijections corresponding to the following simple generating function identities. Let $B(z)=\dfrac{1}{\sqrt{1-4z}}$ and $C(z)=\dfrac{1-...
Alexander Burstein's user avatar
2 votes
2 answers
894 views

Proving generating functions equality

What do you use to prove the following equality (and possibly more general ones of the kind)? \begin{align*}\sum_{r,s,t} \frac{q^{r^2+rs+s^2+st+t^2}}{(q)_r (q)_s (q)_t} z_1^{r+s} z_2^{s+t} = \sum_{a,...
Andrew's user avatar
  • 21
3 votes
2 answers
656 views

p-adic poly-Bernoulli numbers

We can define p-adic Bernoulli polynomials by using q-integral on $\mathbb{Z}_p$ and Taekyun Kim's method. But how can we define p-adic poly-Bernoulli numbers and polynomials by using integral on $\...
user avatar
6 votes
2 answers
912 views

Closed form or/and asymptotics of a hypergeometric sum

Dear mathematicians, I am a computer scientist wandering in the deep sea of combinatorics and asymptotics to pursue a recent interest in average case analysis of algorithms. In doing so, I designed ...
Christian Rinderknecht's user avatar
5 votes
0 answers
345 views

When does a triangle of numbers have a zero row sum?

Suppose we have a triangle of numbers defined by the recurrence relation $$\left| n \atop k \right| = f(n,k) \left| n-1 \atop k \right| +g(n,k) \left| n-1 \atop k-1 \right| + [n=k=0],$$ for some ...
Mike Spivey's user avatar
  • 3,283
2 votes
1 answer
751 views

Transfinite sums related to a sequence

Given a sequence $S$ indexed by the finite ordinals, a limit ordinal $\alpha$, and $k \in \mathbb{N}$, define $S_{\alpha+k}$(the extension of $S$ to $\alpha+k$) to be the sum over the products of all $...
Michael Burge's user avatar
10 votes
2 answers
2k views

Expressions involving Eulerian numbers of the second kind: trying to show $\sum_{m=0}^{n} (-1)^m(m)m!(2n-m-2)!\left\langle\left\langle n\atop m\right\rangle\right\rangle\neq0$ for even $n$.

Considering the success of a previous question involving Eulerian numbers, I thought I might throw this question into the mix. It comes from some localization computations in GW theory, but in this ...
Steffen Marcus's user avatar
26 votes
3 answers
4k views

Is the sum $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0?$

I am trying to prove $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$. This inequality has been verified by computer for $k\le40$. Some clues that might work (kindly provided by ...
Alexandra Seceleanu's user avatar