# Questions tagged [q-analogs]

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70
questions

**5**

votes

**0**answers

108 views

### A recursion which defines polynomials with integer coefficients?

Let $[n]=1+q+\dots+q^{n-1}$ and $u(n)=\prod_{j=1}^n \gcd([j],[n])$.
Define
$$r(n)=\sum_{d|n,d>1}{(-1)^d \frac{u(n)}{du(\frac{n}{d})^d}r\Big(\frac{n}{d}\Big)^d}+\frac{(1-q)^{n-1}u(n)}{n[n]}$$ with $...

**6**

votes

**1**answer

237 views

### Relationship between $q$-Weyl dimension formula and $q$-analog of weight multiplicity?

$\DeclareMathOperator\dim{dim}$For a dominant (integral) weight $\lambda$ and any (integral) weight $\mu$ of a simple Lie algebra $\mathfrak{g}$, Lusztig's $q$-analog of weight multiplicty $K_{\lambda,...

**7**

votes

**7**answers

531 views

### Important combinatorial and algebraic interpretations of the coefficients in the polynomial $[n]!_q = (1+q)(1+q+q^2) \ldots (1+q+\cdots + q^{n-1})$

What are some important combinatorial and algebraic interpretations of the coefficients in the polynomial
$$[n]!_q = (1+q)(1+q+q^2) \ldots (1+q+\cdots + q^{n-1})?$$
As motivation, I will give ...

**11**

votes

**1**answer

330 views

### $q$-analogs of total positivity

A real matrix $M$ is called totally positive if all of its minors are positive; these matrices have been extensively studied, and there are generalizations to other Lie types, for example by Lusztig.
...

**21**

votes

**2**answers

666 views

### A q-rious identity

Let $[x]_q=\frac{1-q^x}{1-q}$, $[n]_q!=[1]_q[2]_q\cdots[n]_q$ and ${\binom{x}{n}}_{q}=\frac{[x]_q[x-1]_q\cdots[x-n+1]_q }{[n]_q!}$.
Computer experiments suggest that
$$\det \left(q^\binom{i-j}{2}\...

**6**

votes

**1**answer

213 views

### Lusztig's $q$-analog of weight multiplicity with product formula

For partitions $\lambda, \mu \vdash n$, the Kostka-Foulkes polynomial $K_{\lambda,\mu}(q)$, a $q$-analog of the Kostka coefficient $K_{\lambda,\mu}$, has a combinatorial description, due to Lascoux ...

**8**

votes

**1**answer

206 views

### Prominent examples of $q$-analogs without known cyclic sieving

The cyclic sieving phenomenon is nicely summarized in the following AMS Notices "What is...?" article: https://www.ams.org/notices/201402/rnoti-p169.pdf.
In that article, Reiner, Stanton, and White ...

**8**

votes

**1**answer

243 views

### Product of $q$-analogues

Background
Recall that the $q$-analogue $[n]_q\in\mathbb Z[q]$ of a natural number $n\in\mathbb N$ is defined as
$$ [n]_q := \frac{q^n -1}{q-1}$$
the idea being that formulas involving $q$ will ...

**6**

votes

**0**answers

101 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 ...

**4**

votes

**0**answers

104 views

### Positivity of q-analogs of central binomial coefficients?

With the usual $q-$notations
$[n]_q=1+q+\cdots+q^{n-1}=\frac{\,\,1-q^n}{1-q},$
$[n]_q!=[1]_q[2]_q\cdots[n]_q$ and
$\binom{n}k_q=\frac{[n]_q!}{[k]_q!\cdot[n-k]_q!}$
let
$$b(n,k,r,q)=\det\left(q^{r\...

**1**

vote

**2**answers

181 views

### $q$-factorial coefficient asymptotics

Consider the $[n]!_q = \prod\limits_{k = 1}^{n} \frac{q^k - 1}{q - 1} = \sum\limits_{k = 0}^{\binom n 2} c_k q^k$ and let $\{f_n\}_{n \in \mathbb{N}}$ be the sequence of the functions on $[0; 1]$ ...

**5**

votes

**0**answers

120 views

### Conjecture for a certain Cauchy-type determinant

Given the Cauchy-like matrix
$$
\mathbf X_M(q) = \left[ \frac{2}{\pi} \frac{
\Gamma\!\left(m - \frac{1}{2} \right)\Gamma\!\left(n + \frac{1}{2} \right)
}{
\Gamma(m)\,\Gamma(n)
}
\frac{m-\frac{3}{4}}
{\...

**2**

votes

**1**answer

147 views

### Major index generating polynomial for border-strip tableaux

The Question in its original form has been answered, but there is a follow-up, see the end of the post.
A border-strip is a skew Young diagram that does not contain a $2 \times 2$-box. A border-strip ...

**6**

votes

**0**answers

175 views

### A curious $q$-identity

Let $[x]_{q}=\frac{1-q^x}{1-q}$ and $\binom{x}{n}_{q}$ denote a $q$-binomial coefficient.
Let $A_n(x,q)$ be the $n\times n $ matrix with entries $$q^\binom{i-j}{2}\binom{i+j+x}{i-j+1}_{q},$$ $0 \le i,...

**6**

votes

**0**answers

111 views

### Irreducibility of q-factorial plus 1

Let $q$ be a formal variable and for every positive integer $n$ let
$$[n]_q! = 1 (1 + q)(1 + q + q^2) \cdots (1 + q + \cdots + q^{n-1})$$
be the $q$-factorial.
Is it true that $[n!]_q + 1$ is an ...

**8**

votes

**0**answers

185 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 \...

**2**

votes

**2**answers

259 views

### What partial sum formulae exist for this basic hypergeometric series?

I've run into:
$$\sum_{x=1}^{\infty} {x^a\over 1-q^{x}}, \ s.t.\ q\in \mathbb N>1 \ or \ q\in (0, 1),\ a \in \mathbb N$$
I am interested mostly in the cases where $a = 1$ or $ a = 2$
Things I'...

**12**

votes

**2**answers

351 views

### $q$ as a prime power and a root of unity

The number of points on the $(n-1)$-dimensional projective space $P^{n-1}(\mathbb{F}_q)$ over a finite field $\mathbb{F}_q$ is the $q$-integer
$$[n]_q := \frac{q^n-1}{q-1}.$$
In analogy, the number of ...

**15**

votes

**1**answer

506 views

### Schur-Weyl duality and q-symmetric functions

Disclaimer: I'm far from an expert on any of the topics of this question. I apologize in advance for any horrible mistakes and/or inaccuracies I have made and I hope that the spirit of the question ...

**10**

votes

**4**answers

642 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 ...

**9**

votes

**0**answers

162 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 ...

**14**

votes

**1**answer

511 views

### Is there a lift of the q-Vandermonde identity to some geometric (motivic) identity for Grassmannians over $F_q$?

The q-Vandermonde identity reads:
$$ \binom{m + n}{k}_{\!\!q} =\sum_{j} \binom{m}{k - j}_{\!\!q} \binom{n}{j}_{\!\!q} q^{j(m-k+j)} $$
The q-binomial coefficients:
$$ \binom{ a }{ b}_{\!\!q} $$
...

**30**

votes

**1**answer

1k views

### Mysterious symmetry - in search for a bijection

I have a mysterious symmetry that I have not managed to prove.
First some definitions (see picture below)
Fix a partition that fit in a staircase shape with $n$ rows.
There are $Catalan(n)$ such ...

**5**

votes

**1**answer

307 views

### $q$-analog of an integral from quantum field theory?

This question has been completely reformulated and a new property for the function $f_q$ has been added due to a series of helpful comments by fedja.
Consider the integral from quantum field theory ...

**6**

votes

**1**answer

277 views

### Inequality for functions on [0,1], continued

Let $0<a<1,\; \psi_a(x)=\displaystyle \prod_{j=0}^\infty (1-a^jx).$ For each $ k\in \mathbb{N},$ set
$$f_k(a;x):=\frac{x^k}{(1-a)(1-a^2)\dots (1-a^k)}\,\psi_a(x).$$
Question. Is it true that, ...

**3**

votes

**0**answers

86 views

### Does the Riemann characterization of the hypergeometric function have a q-analog?

This question is inspired by another recent one here, Characterization of the hypergeometric function. The latter is about the classical result of Riemann characterizing the hypergeometric functions ...

**6**

votes

**1**answer

360 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, $q = \exp(2\pi i r/N)$?

**5**

votes

**2**answers

538 views

### Some curious Hankel determinants

Let $f(n,q)=\prod_{j=1}^na(q^j)$ for a polynomial $a(q)$ and let $d(n)=\det(f(i+j,q))_{i,j=0}^n$ be its Hankel determinant.
Computer experiments suggest that
$$\lim_{q\to1}\frac{d(n)}{(q-1)^\binom{n+...

**12**

votes

**1**answer

175 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 ...

**2**

votes

**1**answer

100 views

### Does this q-analogue have a nice closed form? [closed]

Let $[n]_q=1+q+\cdots+q^{n-1}$. Is there a nice closed form of $\sum_{s=1}^i[s]_{q}$? One would expect that the answer will be some q-analog of $\frac{i(i+1)}{2}$, since $\sum_{s=1}^i s=\frac{i(i+1)...

**13**

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 ...

**8**

votes

**3**answers

369 views

### A not quite theta not quite basic hypergeometric function

The study of matrix quantum group coactions on the noncommutative disk algebra turns up the following series, which is a $q$-deformation of the negative binomial series, for integer $t\ge 0$, complex $...

**5**

votes

**0**answers

232 views

### Is a basic hypergeometric function ${}_2\phi_1(a, b; c; q, z)$ a meromorphic function in $z$?

Here a basic hypergeometric function is the analytic continuation of the basic hypergeometric series (or called the $q$-hypergeometric series)
$$
{}_2\phi_1(a, b; c; q, z) = \sum^{\infty}_{n = 0} \...

**6**

votes

**0**answers

221 views

### $q$-crystals - is there such a thing?

There are several important facts that I first heard about here on MO. One of the most enlightening of these is that $\mathscr D$-modules on a scheme $X$ may be viewed as sheaves on the groupoid of ...

**3**

votes

**1**answer

148 views

### Does $\sum_{n=-\infty}^\infty (bq^n,p/aq^n;p)_\infty z^n q^{n(n-1)/2}$ have a closed form?

The formula
$$
\small\sum_{n=-\infty}^\infty (bq^n,p/aq^n;p)_\infty
z^n q^{n(n-1)/2}=\frac{(-z,-q/z;q)_\infty}{\ln\frac{1}{q}}\int\limits_0^\infty\frac{\left(bt/z,pz/at;p\right)_\infty}{\left(-t,-...

**11**

votes

**2**answers

507 views

### Does $q$-Catalan number count subspaces?

Consider the $n$-element subsets $\{a_1<a_2<\cdots <a_n\}$ of $\{1,\ldots ,2n\}$ satisfying $a_i\geq 2i$ for all $i=1,\ldots ,n$. The number of such subsets is given by $${2n\choose n}-{2n\...

**3**

votes

**1**answer

217 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-...

**13**

votes

**2**answers

636 views

### A conjecture about algebraic values of $(-q;\,-q)_\infty/(q;\,q)_\infty$

Recall that $(a;\,q)_\infty$ is the $q$-Pochhammer symbol:
$$(a;\,q)_\infty=\prod_{n=0}^\infty(1-a \, q^n).\tag1$$
Its important special case $(q;\,q)_\infty=\prod_{n=1}^\infty(1-q^n)$ is sometimes ...

**4**

votes

**0**answers

112 views

### A $q-$binomial identity related to $q-$Narayana polynomials of type B

Denote by $ {n\brack {k}}$ a $q-$binomial coefficient.
Let ${D_{n,k}}(t,q) = \sum\limits_{j = 0}^{n - k} {{q^{{j^2} + kj}}}{n\brack {j}}{n\brack {k+j}}t^j $
and
${R_n}(x,t,q) = \sum\limits_{k = 0}...

**10**

votes

**3**answers

653 views

### Is this a q-count of Alternating Sign Matrices?

The number of Alternating Sign Matrices of size $n$ is well known to be
$\prod_{k=0}^{n-1}\frac{(3k+1)!}{(n+k)!}$. Is it known whether the naive q-analog expression
$$\prod_{k=0}^{n-1}\frac{[3k+1]_q!}{...

**8**

votes

**1**answer

388 views

### q-analog of a combinatorial identity involving binomial coefficients

Using, e.g., properties of iterated finite differences it is easy to show that for any pair of integers $n$ and $m$ with $n>\!>m$ one has the identity
$$
\sum_{k=0}^m(-1)^{k-m} {n-k\choose m}{m\...

**6**

votes

**0**answers

255 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 ...

**12**

votes

**1**answer

334 views

### Multiplicative infinitesimals in q-analogs?

Risking to be downvoted, here is a very lightweight question.
In various fields - say, algebraic geometry, nonstandard analysis, synthetic differential geometry - infinitely small quantities, i. e. ...

**12**

votes

**1**answer

813 views

### Generating function for certain partitions (with a restriction on the Durfee square)

First of all my apologies if this question is well known or obvious: this is not in my area of research.
Let $T(x)=\sum_{n=0}^\infty t_nx^n$, where $t_n$ is the number of partitions $\lambda$ of $n$ ...

**8**

votes

**1**answer

255 views

### notation for $(a-b)(a-qb)\dots (a-q^{n-1}b)$

I wonder whether there is a notation for such thing, which I denote $[a;b]_q^n$ for a moment:
$$
[a;b]_q^n:=(a-b)(a-qb)\dots (a-q^{n-1}b)=a^n(b/a;q)_n,
$$
this last equation uses $q$-Pochhammer symbol ...

**4**

votes

**2**answers

222 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}=...

**8**

votes

**1**answer

378 views

### q-Integer-valued polynomials

For $n \in \mathbb{Z}_{\geq 0}$, let $[n]_q := (1-q^n)/(1-q) = (1+q+...+q^{n-1})$ as is customary, with $[0]_q=0$.
Let $R$ be the subring of $\mathbb{Q}(q)[x]$ consisting of all $f$ such that $f([n]...

**8**

votes

**0**answers

541 views

### What is the $q$-analog of $\Gamma(z)\Gamma(1-z)=\frac\pi{\sin(\pi z)}$?

I would expect the $q$-Gamma function to have the property which would be the $q$-analog of the Euler reflection formula from my question title.
More concretely: $\Gamma(z)$ has simple poles at ...

**13**

votes

**2**answers

1k views

### Is there a $q$-L'Hospital's Rule?

Let $\binom{n}{j}_q$ be a $q$-binomial coefficient and $(x;q)_n = (1-x)(1-qx)\cdots(1-q^{n-1}x).$
Consider the sum $$f(n,m,r,k)= \sum\limits_{j = 0}^{2n} {( - 1)}^{ j}q^{mj^2+rj}
\binom{2n}{j}_{q^k}$...

**8**

votes

**1**answer

805 views

### Counting subspaces

We are given the finite vector space $V = V(n,p) = \mathbb{F}_p^n$ and two fixed subspaces $W_1, W_2 \subseteq V$ of dimensions $m_1$, $m_2$ respectively. Suppose
that the dimension of $W_1 \cap W_2$ ...