# Questions tagged [combinatorial-identities]

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151
questions

3
votes

3
answers

647
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### Ordinary partitions vs partitions into odd parts

Let $\mathcal{P}(n)$ be the set of all unrestricted partitions of $n$ while $\mathcal{O}(n)$ stand for the set of all partitions of $n$ into odd parts. We adopt the power notation for partitions $\...

6
votes

1
answer

310
views

### A summation involving fraction of binomial coefficients

I need to prove the following statement.
Let $ n, g, m, a ,t$ be integers. Prove that the following statement is true for all $ n \geq g(1+2m)+1 $, $ g\geq 2t $, $ m\geq t $, $ 0\leq a <t $, and $ ...

7
votes

2
answers

501
views

### An identity involving polylogarithms

Recall that
$$\mathrm{Li}_2(x):=\sum_{n=1}^\infty\frac{x^n}{n^2}.$$
I have found the following identity:
\begin{equation}\begin{aligned}&\mathrm{Li}_2\left(\frac{-1-\sqrt{-7}}4\right)+\mathrm{Li}...

4
votes

0
answers

89
views

### "Convolving" a general Catalan with classical Catalan

Consider what is sometimes known as generalized Catalan sequence
$$\mathcal{{\color{red}C}}_{a,b}:=\frac{2b+1}{a+b+1}\binom{2a}{a+b}.$$
Observe that $\mathcal{{\color{red}C}}_{n,0}$ reduces to the ...

4
votes

0
answers

196
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+\...

0
votes

0
answers

186
views

### Looking for a combinatorial proof of an identity

I've come up with an interesting combinatorial identity (thanks to P. Belmans who precomputed the numbers and pointed out to me that they correspond to OEIS A002697):
$$
\sum_{i=0}^{n-1}\binom{n+1-i}{...

3
votes

1
answer

194
views

### Seeking for a combinatorial argument for partition identities

Given an integer partition $\lambda$, introduce the following quantities:
\begin{align*}
c(\lambda)&=\sum_{i\geq1}\left\lceil\frac{\lambda_i}2\right\rceil, \qquad c_o(\lambda)=\sum_{i\geq1}\left\...

10
votes

2
answers

1k
views

### Proving an identity about Catalan numbers

$$C_{n} = \sum_{i=1}^n (-1)^{i-1} \binom{n-i+1}{i} C_{n-i}$$
Are there any good combinatorial proofs or algebraic proofs of this?

0
votes

0
answers

211
views

### An alternating sum involving a product of binomial coefficients

I encountered the sum below, where $c_{1}$, $c_{2}$, $c_{3}$, $c_{4}$ and $d$ are some given positive constants. Does anyone have an idea how to simplify it?
$$
\sum\limits_{k=1}^{d} \frac{(-1)^{k-1}k}...

7
votes

1
answer

235
views

### A reference for a sum found in Gould's Combinatorial Identities book

On p. 49 in Gould's book Combinatorial Identities, the author states that the sum $$\sum_{k=0}^{n-1}(-1)^k\binom{n}{k}\binom{2n}{2k}^{-1}$$ "... arises naturally in a statistical problem; it ...

2
votes

1
answer

183
views

### A Vandermonde like determinant with exponentials

Let $n\geq m$ be non negative integers, and consider a list of $(n+m+1)$ distinct numbers (complex or real). I am interested in getting a closed form formula for the following determinant: $\det\left[\...

10
votes

0
answers

428
views

### New series for $\zeta(5)$ involving second-order harmonic numbers

In 1997 T. Amdeberhan and D. Zeilberger proved that
$$\sum_{k=1}^\infty\frac{(-1)^k(205k^2-160k+32)}{k^5\binom{2k}k^5}=-2\zeta(3).\tag{1}$$
In 2008 J. Guillera obtained that
$$\sum_{k=1}^\infty\frac{(...

9
votes

1
answer

312
views

### Series for $\frac{\log m}{\pi}$ with summands involving harmonic numbers

The classical rational Ramanujan-type series for $1/\pi$ have the following four forms:
\begin{align}\sum_{k=0}^\infty(ak+b)\frac{\binom{2k}k^3}{m^k}&=\frac{c}{\pi},\label{1}\tag{1}
\\\sum_{k=0}^\...

4
votes

0
answers

195
views

### Curious double sum identity

The following identity seems to hold for $a>1$ :
$$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{ \frac{a^m}{m}\left( \frac{a^m}{m}+ \frac{a^n}{n} \right) } = \frac{a^2}{2(a-1)^4}$$
I've tested ...

1
vote

1
answer

146
views

### Curious identity involving the number of perfect matchings of the complete graph

Can you prove (preferably combinatorially) the following identity for the total number of perfect matchings of the complete graph $K_{2n}$, where the edges in the matching are ordered, i.e., $\binom{...

8
votes

1
answer

519
views

### Trivial (?) product/series expansions for sine and cosine

In an old paper of Glaisher, I find the following formulas:
$$\dfrac{\sin(\pi x)}{\pi x}=1-\dfrac{x^2}{1^2}-\dfrac{x^2(1^2-x^2)}{(1.2)^2}-\dfrac{x^2(1^2-x^2)(2^2-x^2)}{(1.2.3)^2}-\cdots$$
$$\cos(\pi x/...

12
votes

0
answers

375
views

### $q$-analogue of the multinomial theorem?

The $q$-binomial theorem states that
$$
\prod_{k=0}^{n-1}(1+q^kt) = \sum_{k=0}^n q^{\binom k2}{n\brack k}_q t^k.
$$
This identity is a $q$-analogue of the binomial theorem
$$
(1+t)^n = \sum_{k=0}^n \...

1
vote

1
answer

206
views

### Gaussian at $q=\pm1$, log-concave polynomials, Catalan numbers

Let $[n]_q!=\prod_{j=1}^n\frac{1-q^j}{1-q}$ with $[0]_q!:=1$ and the Gaussian polynomials $\binom{n}k_q=\frac{[n]_q!}{[k]_q!\,\cdot\,[n-k]_q!}$. Adopt the convention that $\binom{n}k_q=0$ whenever $k&...

3
votes

2
answers

412
views

### Combinatorial identity concerning integral matrices with prescribed row sums and column sums

How to prove the following identity?
Let $r = (r_1, r_2, \ldots, r_d)$ and $c = (c_1, c_2, \ldots, c_d)$ be sequences of natural numbers such that $s = r_1 + r_2 + \cdots + r_d = c_1 + c_2 + \ldots + ...

2
votes

1
answer

189
views

### $q$-binomial sum, slightly

Recall that $[n]_{q}!=\prod_{j=1}^n\frac{1-q^{j}}{1-q}$ and $\binom{n}k_{q}=\frac{[n]_{q}!}{[k]_{q}![n-k]_{q}!}$. Then the $q$-binomial theorem states
$$\sum_{k=0}^n\binom{n}k_qq^{\binom{k}2}=\prod_{k=...

5
votes

1
answer

285
views

### A conjectural permanent identity

Let $n>1$ be an integer, and let $\zeta$ be a primitive $n$th root of unity. By $(3.4)$ of arXiv:2206.02589, $1$ and those $n+1-2s\ (s=1,\ldots,n-1)$ are all the eigenvalues of the matrix $M=[m_{jk}...

3
votes

1
answer

347
views

### Is this combinatorial identity known? (of interest for random matrix theory)

While playing around with random matrices and I arrived at a different formula for the mean of the limiting normal distribution for a spectral CLT for sample covariance matrices. More precisely I have ...

2
votes

1
answer

468
views

### Integer eigenvalues of a class of matrices inspired by Prof. Zhi-Wei Sun's conjecture

Theorem: Let $n>1$ be an odd number and $\zeta$ a primitive $n$-th root of unity. Then
\begin{eqnarray}
&&\sum_{\tau\in D(n-1)}\mathrm{sign}(\tau)\prod_{j=1}^{n-1}\frac{1}{1-\zeta^{j-\tau(j)...

1
vote

1
answer

206
views

### A conjectural identity involving infinite series

Recently I formulated the following curious conjecture based on my computation.
Conjecture. For all $|x|>1$, we have the identity
$$\sum_{k=0}^\infty\frac{\sum_{j=0}^{k}\binom{2k+1}{2j}(1-x)^jx^{k-...

-1
votes

1
answer

321
views

### Could you please confirm or deny two identities involving weighted Stirling numbers of the second kind?

In the paper [1] below, among other things, Carlitz introduced weighted Stirling numbers of the second kind $R(n,k,r)$. He also proved that the numbers $R(n,k,r)$ can be generated by
\begin{equation*}%...

2
votes

2
answers

245
views

### Ask for a proof of an identity involving the product of two Bernoulli numbers

It is well known that the Bernoulli numbers $B_{k}$ for $k\in\{0,1,2,\dotsc\}$ can be generated by
\begin{equation*}
\frac{z}{\textrm{e}^z-1}=\sum_{k=0}^\infty B_k\frac{z^k}{k!}=1-\frac{z}2+\sum_{k=1}^...

15
votes

2
answers

935
views

### A rather curious identity on sums over triple binomial terms

While exploring the Baxter sequences from my earlier MO post, I obtained a rather curious identity (not listed on OEIS either). I usually try to employ the Wilf-Zeilberger (WZ) algorithm to justify ...

9
votes

2
answers

407
views

### Identity involving a quadratic term inside the Pochhammer symbol

This identity came up in my research:
$$
\sum_{m=1}^n m^2 \frac{(\frac{xy}n + m-1)_{2m-1} (n+m-1)_{2m-1}}{(x+m)_{2m+1} (y+m)_{2m+1}} = \frac{n^2}{(x^2-n^2) (y^2 - n^2)}.
$$
Here $n$ is a fixed ...

3
votes

2
answers

402
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-...

4
votes

1
answer

134
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}\...

12
votes

3
answers

833
views

### Set partitions and permanents

Let $a(n)=$ Number of ordered set partitions of $[n]$ such that the smallest element of each block is odd.
...

6
votes

1
answer

485
views

### A novel identity connecting permanents to Bernoulli numbers

For a matrix $[a_{j,k}]_{1\le j,k\le n}$ over a field, its permanent is defined by
$$\mathrm{per}[a_{j,k}]_{1\le j,k\le n}:=\sum_{\pi\in S_n}\prod_{j=1}^n a_{j,\pi(j)}.$$
In a recent preprint of mine, ...

10
votes

1
answer

511
views

### Identities involving derangements and roots of unity

For a positive integer $n$, a derangement of $\{1,\ldots,n\}$ is a permutation $\tau$ of $\{1,\ldots,n\}$ with $\tau(j)\not=j$ for all $j=1,\ldots,n$. For convenience, we let $D(n)$ denote the set of ...

9
votes

1
answer

634
views

### Permanent identities

The permanent $\mathrm{per}(A)$ of a matrix $A$ of size $n\times n$ is defined to be:
$$\mathrm{per}(A)=\sum_{\tau\in S_n}\prod_{j=1}^na_{j,\tau(j)}.$$
Let
$$A=\left[\tan\pi\frac{j+k}n\right]_{1\le j,...

5
votes

3
answers

568
views

### How to prove the combinatorial identity $\sum_{k=\ell}^{n}\binom{2n-k-1}{n-1}k2^k=2^\ell n\binom{2n-\ell}{n}$ for $n\ge\ell\ge0$?

With the aid of the simple identity
\begin{equation*}
\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^{k}}=2^n
\end{equation*}
in Item (1.79) on page 35 of the monograph
R. Sprugnoli, Riordan Array Proofs of ...

3
votes

1
answer

167
views

### Is there a $q$-analogue to Shapiro's convolution identity?

Let $C_n=\frac1{n+1}\binom{2n}n$ denote the Catalan numbers.
This question is motivated by the (unanswered) MO post by Alexander Burstein and my own (answered by Fedor Petrov) MO post.
Specifically, ...

7
votes

1
answer

292
views

### Looking for a $q$-analogue of a binomial identity

The following identity is well-known and there are a few proofs to it (see Bijective proof problems, by R P Stanley, for this and similar formulae):
$$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}=4^n \...

10
votes

0
answers

333
views

### A bijective proof for the odd companion to Shapiro's Catalan convolution

Shapiro's Catalan convolution is the following formula (where $C_n$ is the $n$th Catalan number):
$$
\sum_{k=0}^{n}{C_{2k}C_{2(n-k)}}=4^nC_n.
$$
In other words, letting $C(z)=\sum_{n=0}^{\infty}{C_nz^...

6
votes

2
answers

208
views

### A convolution-type identity for the "major index"

For a permutation $\pi\in\frak{S}_n$, define the number of descents of $\pi$ as $$\text{des}(\pi)=\vert\{i: \pi(i)>\pi(i+1)\}\vert.$$
The following is a well-known (and interesting) identity:
$$\...

3
votes

2
answers

68
views

### Optimal scaling of Lipschitz estimates in generalized geometric series

If we did not know it before, then wikipedia teaches us the generalized geometric series
$$\sum_{n \ge 0} \binom{n+k}{n} (1-\mu)^{n} \mu^k = \frac{1}{\mu}.$$
We can then study for $0 <\varepsilon &...

6
votes

1
answer

153
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}) \...

2
votes

1
answer

250
views

### Evaluations of three series involving binomial coefficients

Question. How to prove the following three identities?
\begin{align}\sum_{k=1}^\infty\frac1{k(-2)^k\binom{2k}k}\left(\frac1{k+1}+\ldots+\frac1{2k}\right)=\frac{\log^22}3-\frac{\pi^2}{36},\tag{1}
\end{...

3
votes

2
answers

240
views

### Is there a combinatorial reason for variable-independence of this binomial-coefficient identity?

Consider the following identity
$$\sum_{n=0}^{R-t}\binom{n+\ell}n\binom{R-\ell-n}{R-t-n}=\binom{R+1}{t+1}.\tag1$$
It is relatively easy to give an algebraic or mechanical proof of (1). But, I like to ...

7
votes

1
answer

190
views

### Reference for permanent integral identity

$\DeclareMathOperator\perm{perm}\DeclareMathOperator\diag{diag}$Using MacMahon's master theorem, the properties of complex gaussian integrals, and Cauchy's integral theorem one can show that the ...

18
votes

0
answers

732
views

### Two curious series for $1/\pi$

On Jan. 18, 2012 I conjectured that for any prime $p>3$ we have
$$R_p^2\equiv\frac1{10}\left(512\left(\frac{10}p\right)-27\left(\frac{-15}p\right)-475\right)\pmod p,$$
where $(\frac{\cdot}p)$ ...

3
votes

1
answer

416
views

### How to prove this combinatorial identity?

If $n \in \mathbb N \setminus \{0\}$ and $x,y,z \in \mathbb R$ such that $x+y+z=n-1$, show that
$$\dfrac{(-4)^n}{\binom{2x}{n}}\sum_{r+s=n,r,s\in Z}\dfrac{\binom{y}{r}\binom{y-a}{r}\binom{z}{s}\binom{...

6
votes

5
answers

838
views

### Combinatorial proof of Catalan's identity

Consider the problem of tiling a board of length $n$ with squares of size $1×1$ and dominoes of size $1×2$, Let's denote $f_n$ to be the number of ways to tile this so-called ($n$)-board.Then $f_n=F_{...

3
votes

1
answer

423
views

### An identity for polynomials over partitions

Given an integer partition $\lambda=(\lambda_1,\dots,\lambda_{\ell(\lambda)})$ of $n$ where $\ell(\lambda)$ is the length of $\lambda$, associate its conjugate partition $\lambda'$. Denote by $\lambda'...

14
votes

2
answers

482
views

### Curious identity between the two kinds of Chebyshev polynomials

I have found, by accident, an identity that relates a sum of Chebyshev polynomials of the first kind to a Chebyshev polynomial of the second kind. It goes as follows:
Given an integer partition of $n$...

2
votes

0
answers

194
views

### For human proofs of two novel combinatorial identities

For $n=0,1,2,\ldots$, let us define the polynomial
$$S_n(x):=\sum_{k=0}^n\binom{x/2}k\binom{(x-1)/2}k\binom{-(x+1)/2}{n-k}\binom{-(x+2)/2}{n-k}.$$
Such polynomials occur in some series for $1/\pi$ ...