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12 votes
3 answers
1k views

A "quantum" identity: in search of a proof -Part II

As usual, denote $[n]_q=1+q+\cdots+q^{n-1}=\frac{\,\,1-q^n}{1-q}$ and $[n]_q!=[1]_q[2]_q\cdots[n]_q$. Furthermore, we write $$\binom{n}k_q=\frac{[n]_q!}{[k]_q!\cdot[n-k]_q!}.$$ As a follow up on this ...
T. Amdeberhan's user avatar
12 votes
2 answers
1k views

An interesting identity: in search of a proof -Part I

I like the following binomial identity in that the RHS extracts the indeterminate $w$ from the LHS. Question. Can you show that $$\sum_{k=0}^n\binom{x+kw}k\binom{y-kw}{n-k}=\sum_{k=0}^n\binom{x+y-...
T. Amdeberhan's user avatar
12 votes
1 answer
730 views

Two remarkable weighted sums over binary words

This question builds off of the previous MO question Number of collinear ways to fill a grid. Let $A(m,n)$ denote the set of binary words $\alpha=(\alpha_1,\alpha_2,\ldots,\alpha_{m+n-2})$ consisting ...
Sam Hopkins's user avatar
  • 24.2k
11 votes
3 answers
557 views

In search of a $q$-analogue of a Catalan identity

Let $C_n=\frac1{n+1}\binom{2n}n$ be the all-familiar Catalan numbers. Then, the following identity has received enough attention in the literature (for example, Lagrange Inversion: When and How): \...
T. Amdeberhan's user avatar
4 votes
1 answer
376 views

Counting permutations with a fixed number of descents and an extra condition

I am computing the volumes of certain polytopes and it turns out that knowing a "closed formula" for the following number would help a lot. Determine the number of permutations $\sigma\in \...
Luis Ferroni's user avatar
  • 1,889
3 votes
3 answers
396 views

Chebyshev polynomials and ballot numbers

I have asked this question a short time ago on mathstackexchange, but it has already fallen into the abyss of answered and uncommented questions. So I take the risk to ask it on mathoverflow. Playing ...
Libli's user avatar
  • 7,300
3 votes
1 answer
392 views

A combinatorial identity involving binomial coefficients

When I was reading an article by CHUN-GANG JI (A SIMPLE PROOF OF A CURIOUS CONGRUENCE BY ZHAO), he mentioned in the acknowledgement the following identity $$\sum_{i+j+k=p,\text{ } i,j,k\gt 0}{p\choose ...
wkmath's user avatar
  • 53
3 votes
1 answer
253 views

What is the value of this sum involving q-binomials?

Let $n\ge 2r$ be positive integers. Is there a closed form for following finite summation involving in q-binomial coefficients $$\sum_{s=0}^r(-1)^sq^{\frac{s(s+1)}{2}}{n-2r+s\brack n-2r}_q{n\brack r-...
Bumblebee's user avatar
  • 1,093
3 votes
0 answers
274 views

Inequalities for Motzkin polynomials

Let us denote by $M_{n}(t)$ the $n$-th Motzkin polynomial. It is defined by $M_1(t) = M_2(t) = 1$ and $$ M_{n}(t) = \sum_{i=0}^{\lfloor n/2\rfloor } \frac{1}{n-1-i} \binom{n-1-i}{i} \binom{n-1}{i+1} t^...
Luis Ferroni's user avatar
  • 1,889
2 votes
3 answers
742 views

Asking for a proof for a sum of products of binomials: an "interesting" identity?

The following identity must have received alternative proofs, including a combinatorial argument by David Callan as found at Bijections for the Identity $4^n = \sum_{k = 0}^n \binom{2k}k\binom{2(n - k)...
T. Amdeberhan's user avatar