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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
0 answers
631 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 \...
Amritanshu Prasad's user avatar
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
6 votes
0 answers
342 views

What is known about the $q$-analogue of the simplex?

I am interested in the field with one element. I am thus interested in combinatorial interpretations of the Gaussian binomial coefficients. Richard Stanley's "Enumerative combinatorics" mentions ...
Andrius Kulikauskas's user avatar
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
1 answer
186 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, ...
T. Amdeberhan's user avatar
1 vote
1 answer
232 views

Looking for q-analog of derangement anagrams for a word

I have already known QPermutationDerangement: It describes the distribution $$ d_n(q)=\sum_{\sigma \in D_n} q^{\operatorname{maj}(\sigma)} $$ Where we sum over all derangements of an $n$ element set. ...
138 Aspen's user avatar
  • 175