All Questions
Tagged with q-analogs binomial-coefficients
14 questions
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):
\...
- 43.2k
0
votes
0
answers
174
views
3D generalization of Gaussian q-binomial coefficient
It is known that the coefficient of $q^t$ in Gaussian binomial coefficient $\binom{m+n}m_q$ equals the number of permutations of the multiset $\{0^m, 1^n\}$ with $t$ inversions.
Is there a closed ...
- 34.3k
12
votes
5
answers
836
views
A divisibility of q-binomial coefficients combinatorially
Let a and b be coprime positive integers. Then the number a+b divides the binomial coefficient ${a+b \choose a}$. I know how to prove this combinatorially - for example after choosing an ordered set ...
- 8,866
6
votes
0
answers
214
views
Looking for a combinatorial proof for an identity involving $q$-Catalan triangles
Let $C_n=\frac1{n+1}\binom{2n}n$ be the Catalan numbers. Following my earlier post on MO, one fine colleague asked me if there is a $q$-analogue of the identity formed by the so-called Shapiro's ...
- 43.2k
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 ...
- 43.2k
3
votes
0
answers
136
views
A recursion involving binomial coefficients: looking for a q-analog
Let $a_n := \frac{1}{2n+1}\binom{3n}{n}$.
Then it is known that (one can find references in the OEIS for this.)
$$
a_n = \sum_{\substack{i,j,k \geq 0 \\ i+j+k=n-1} } a_i a_j a_k.
$$
Is there a natural ...
- 15.8k
5
votes
1
answer
178
views
A $q$-analogue of a characterization of polynomials by binomial coefficients
Considering the binomial coefficient $\binom{x}{m}$ as a polynomial in $x$, the span of $\binom{x}{0}, \binom{x}{1}, \ldots, \binom{x}{d}$ is exactly the polynomials of degree $\le d$. A closely ...
- 11.2k
12
votes
1
answer
267
views
Total positivity of $q$-Pascal matrix?
A matrix of real numbers is called totally positive if all its minors are non-negative. A well-known example is the Pascal matrix $(\binom{i}{j})$.
Is it true that the minors of the $q$-Pascal matrix ...
- 5,622
6
votes
1
answer
689
views
Q-binomials at roots of unity
As the title says, given a general $q$-binomial $\binom{n}{k}_q$,
is there some general result regarding its value at a root of unity, $q = \exp(2\pi i r/N)$?
- 15.8k
8
votes
0
answers
253
views
q-analog of $(d/dx) \binom{x}{k}$?
It is not hard to find easy formulas for the derivative of the function $\binom{x}{k}$, for instance it is not too hard to see (for $k$ an integer) that
$\frac{d}{dx} \binom{x}{k} = \sum_{i=1}^k \...
- 131
6
votes
0
answers
132
views
Q-analogue of an inequality
Pick integers $b\geq a \geq 0$ and $k\geq j\geq 0$.
It is not super-difficult to prove the inequality
$$
\binom{kb}{ka}^j \geq \binom{jb}{ja}^k.
$$
This is actually quite a nice inequality that was ...
- 15.8k
9
votes
0
answers
192
views
For $q$-analogues of a known curious identity
In 2002 I published the folllowing curious combinatorial identity:
$$(x+m+1)\sum_{i=0}^m(-1)^i\binom{x+y+i}{m-i}\binom{y+2i}i-\sum_{i=0}^m\binom{x+i}{m-i}(-4)^i=(x-m)\binom xm.$$
My original proof is ...
- 15.6k
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
$$\sum_{s=0}^r(-1)^sq^{\frac{s(s+1)}{2}}{n-2r+s\brack n-2r}_q{n\brack r-...
- 1,093
4
votes
2
answers
292
views
How to prove that $\sum_{i=0}^n\frac{(a;q)_i}{(q;q)_i}\frac{(b;q)_{n-i}}{(q;q)_{n-i}}a^{n-i}=\frac{(ab;q)_n}{(q;q)_n}$?
By Cauchy identity, $${}_1\phi_0(a;—;q,z)=\sum_{n\geq0}\frac{(a;q)_n}{(q;q)_n}z^n=\frac{(az;q)_{\infty}}{(z;q)_\infty},\quad|z|<1,|q|<1,$$
we can obtain the $q-$analogue of $(1-z)^{-a}(1-z)^{-b}=...
- 139