Questions tagged [q-analogs]
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101 questions
14
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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:
$$ \binom{ a }{ b}_{\!\!q} $$
...
- 24.9k
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 ...
- 15.8k
5
votes
1
answer
401
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 ...
- 5,624
6
votes
1
answer
340
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, ...
- 783
3
votes
0
answers
106
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 ...
- 18.4k
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
5
votes
2
answers
635
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
$$\lim_{q\to1}\frac{d(n)}{(q-1)^\binom{n+...
- 5,622
12
votes
1
answer
267
views
Total positivity of $q$-Pascal matrix?
A matrix of real numbers is called totally positive if all its minors are non-negative. A well-known example is the Pascal matrix $(\binom{i}{j})$.
Is it true that the minors of the $q$-Pascal matrix ...
- 5,622
2
votes
1
answer
109
views
Does this q-analogue have a nice closed form? [closed]
Let $[n]_q=1+q+\cdots+q^{n-1}$. Is there a nice closed form of $\sum_{s=1}^i[s]_{q}$? One would expect that the answer will be some q-analog of $\frac{i(i+1)}{2}$, since $\sum_{s=1}^i s=\frac{i(i+1)...
- 470
12
votes
3
answers
1k
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A "quantum" identity: in search of a proof -Part II
As usual, denote $[n]_q=1+q+\cdots+q^{n-1}=\frac{\,\,1-q^n}{1-q}$ and $[n]_q!=[1]_q[2]_q\cdots[n]_q$. Furthermore, we write
$$\binom{n}k_q=\frac{[n]_q!}{[k]_q!\cdot[n-k]_q!}.$$
As a follow up on this ...
- 43.2k
8
votes
3
answers
417
views
A not quite theta not quite basic hypergeometric function
The study of matrix quantum group coactions on the noncommutative disk algebra turns up the following series, which is a $q$-deformation of the negative binomial series, for integer $t\ge 0$, complex $...
- 1,143
5
votes
0
answers
388
views
Is a basic hypergeometric function ${}_2\phi_1(a, b; c; q, z)$ a meromorphic function in $z$?
Here a basic hypergeometric function is the analytic continuation of the basic hypergeometric series (or called the $q$-hypergeometric series)
$$
{}_2\phi_1(a, b; c; q, z) = \sum^{\infty}_{n = 0} \...
- 133
7
votes
0
answers
235
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$q$-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 ...
- 18.4k
3
votes
1
answer
1k
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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
$$
\small\sum_{n=-\infty}^\infty (bq^n,p/aq^n;p)_\infty
z^n q^{n(n-1)/2}=\frac{(-z,-q/z;q)_\infty}{\ln\frac{1}{q}}\int\limits_0^\infty\frac{\left(bt/z,pz/at;p\right)_\infty}{\left(-t,-...
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11
votes
2
answers
604
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Does $q$-Catalan number count subspaces?
Consider the $n$-element subsets $\{a_1<a_2<\cdots <a_n\}$ of $\{1,\ldots ,2n\}$ satisfying $a_i\geq 2i$ for all $i=1,\ldots ,n$. The number of such subsets is given by $${2n\choose n}-{2n\...
- 1,305
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
$$\sum_{s=0}^r(-1)^sq^{\frac{s(s+1)}{2}}{n-2r+s\brack n-2r}_q{n\brack r-...
- 1,093
19
votes
1
answer
511
views
"quantum" symmetric plane partitions beget alternating sign matrices?
The "quantum" version qTSPP of the number of totally symmetric plane partitions, contained in the cube $[0,n]^3$, is enumerated by
$$f_n(q):=\prod_{j=1}^n\prod_{k=1}^j\prod_{\ell=1}^k\frac{1-...
- 43.2k
14
votes
2
answers
729
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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 ...
- 6,819
4
votes
0
answers
126
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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}...
- 5,622
11
votes
3
answers
726
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
$$\prod_{k=0}^{n-1}\frac{[3k+1]_q!}{...
- 85.6k
8
votes
1
answer
553
views
q-analog of a combinatorial identity involving binomial coefficients
Using, e.g., properties of iterated finite differences it is easy to show that for any pair of integers $n$ and $m$ with $n>\!>m$ one has the identity
$$
\sum_{k=0}^m(-1)^{k-m} {n-k\choose m}{m\...
- 6,777
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 ...
12
votes
1
answer
351
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. ...
- 18.4k
13
votes
1
answer
982
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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 $n$ ...
- 934
9
votes
1
answer
420
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 ...
- 109k
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
8
votes
1
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497
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]...
- 24.2k
8
votes
0
answers
754
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 ...
- 18.4k
13
votes
2
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1k
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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 $$f(n,m,r,k)= \sum\limits_{j = 0}^{2n} {( - 1)}^{ j}q^{mj^2+rj}
\binom{2n}{j}_{q^k}$...
- 5,622
8
votes
1
answer
1k
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Counting subspaces
We are given the finite vector space $V = V(n,p) = \mathbb{F}_p^n$ and two fixed subspaces $W_1, W_2 \subseteq V$ of dimensions $m_1$, $m_2$ respectively. Suppose
that the dimension of $W_1 \cap W_2$ ...
- 347
22
votes
1
answer
884
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q-Catalan numbers from Grassmannians
In this question by $q$-Catalan numbers I mean the $q$-analog given by the formula $\frac{1}{[n+1]_q}\left[{2n\atop n}\right]_q$. The polynomial $\left[{2n\atop n}\right]_q$ represents the class of ...
- 85.6k
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0
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349
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Some $q-$analogues of $ \sum\limits_{j = - k}^k {{{( - 1)}^{ j}}}\binom{n}{k-j}\binom{n}{k+j}=\binom{n}{k}.$
Let ${\left( {a;q} \right)_n}=\prod\limits_{j = 0}^{n - 1} {(1-{q^j}a} )$ and
let $ {{n}\brack{k}}_q$ denote a $q-$binomial coefficient.
I am interested in $q-$analogues of the identity $ \sum\...
- 5,622
8
votes
5
answers
2k
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Infinite matrix leading eigenvector problem
This question is cross-posted at Math.StackExchange.com.
I'm trying to find the leading eigenvalue and corresponding left and right eigenvectors of the following infinite matrix, for $\lambda>0$:
$$...
- 1,344
12
votes
0
answers
502
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A $q$-analogue of Foulkes' character related to alternating permutations
My paper "Alternating permutations and symmetric functions" at
http://math.mit.edu/~rstan/papers/altenum.pdf enumerates certain
classes of alternating permutations, such as those whose inverse is
...
- 50.8k
3
votes
1
answer
348
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Closed form for a simple hypergeometric $q$ series
I've run across an interesting hypergeometric $q$-series that I feel must have been found before:
$\sum_{n=0}^{\infty}(-1)^n$$\frac{e^{n b y}}{\prod_{k=1}^{n}(e^{\pi k b^2}-e^{\pi k b^{-2}})} = \...
- 151
1
vote
1
answer
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Special values of continuous q - Hermite polynomials
The continuous $q-$Hermite polynomials are defined by
$${H_{n + 1}}(x|q) = 2x{H_n}(x|q) +( {q^n}-1){H_{n - 1}}(x|q)$$
with initial values ${H_{ - 1}}(x|q) = 0$ and ${H_0}(x|q) = 1.$
Cf. e.g. http://...
- 5,622
5
votes
1
answer
2k
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What does the $q$-Catalan Numbers count?
I had completed a paper describing the $q$-Catalan numbers, which is the $q$-analog of the Catalan numbers.
The $n$-th Catalan numbers can be represented by:
$$C_n=\frac{1}{n+1}{2n \choose n}$$
and ...
- 53
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1
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527
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A q,t-extension of Plancherel Measure thru Yang-Mills Theory ?
Buried in the physics paper by Nekrasov and Okounkov, a strange identity is proven:
$$ \prod_{n > 0} (1 - q^n)^{\mu^2-1} = \sum_{\mathbf{k}} q^{|\mathbf{k}|} \prod_{\square \in k} \left( 1 - \frac{\...
- 22.8k
4
votes
1
answer
672
views
Are the following q-Genocchi numbers known?
The sequence of Genocchi numbers
${({G_{2n}})_{n \ge 0}}=$ $(0,1,1,3,17,155,2073,...)$
can be defined by the generating function
$z\frac{{1 - {e^z}}}{{1 + {e^z}}} = \sum {{{( - 1)}^n}{G_{2n}}\frac{...
- 5,622
1
vote
0
answers
413
views
Combinatorial Interpretation of an Extension of Gaussian Polynomials
It is well-known that the Gaussian polynomial (or Gaussian coefficient, q-binomial coefficient) $\binom{n}{k}_q$ counts the number of $k$-dimensional subspaces of an $n$-dimensional vector space over $...
- 443
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5
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2k
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enumerative meaning of natural q-Catalan numbers
Define $[n]=(1-q^n)/(1-q)$ and $[n]!=[1][2][3] \cdots [n]$, so that $[2n]!/[n]![n+1]!$ is a polynomial in $q$ (the most algebraically natural $q$-analogue of the Catalan numbers); what enumerative ...
- 19.7k
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2
answers
1k
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Why are some q-analogues more canonical than others?
It is striking that some q-analogs of functions, operators, identities and especially whole theorems seem quite "canonical", e.g.
the factorial and the q-Gamma function
the basic hypergeometric ...
- 13.4k
6
votes
1
answer
644
views
q-analog of the matrix exponential
I am a fan of the Matrix exponential $\exp(X)$, defined for any complex matrix $X$ by
\begin{equation*}
\exp(X) := \sum_{k \ge 0} \frac{X^k}{k!}.
\end{equation*}
I have a fleeting acquaintance with ...
- 28.6k
12
votes
1
answer
561
views
$(q,x)$-analog of $n!$
While doing some work in geometric representation theory I have come across the following
sequence of polynomials in two variables $(q,x)$ which I would like to denote
by $n!_{q,x}$. For small $n$ ...
- 7,037
4
votes
1
answer
388
views
Taylor expansion of a q-analog of the negative binomial distribution
Given $A,B \in \mathbb{Z}_+$ and $ 0 < t, q< 1$, I'd like to compute the coefficients $c_n(q,A,B)$ in the expansion of the product $$\prod_{i=0}^{A-1} \prod_{j=0}^{B-1} \frac{1}{1-t q^{i+j}} = \...
- 952
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votes
2
answers
3k
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Is there a "quantum" Riemann zeta function?
Occasionally I find myself in a situation where a naive, non-rigorous computation leads me to a divergent sum, like $\sum_{n=1}^\infty n$. In times like these, a standard approach is to guess the ...
- 54.6k
24
votes
6
answers
2k
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Is there a q-analog to the braid group?
The braid group $B_n$ on $n$ strands fits into a short exact sequence of groups:
$$ 1 \longrightarrow P_n \longrightarrow B_n \longrightarrow S_n \longrightarrow 1,$$
where $S_n$ is the symmetric ...
- 5,898
9
votes
1
answer
641
views
A q-analogue of Ramanujan's tau function
There have been a couple of questions on Ramanujan's $\tau$ function.
Lehmer's conjecture for Ramanujan's tau function
The Vanishing of Ramanujan's Function tau(n)
A $q$-analogue is given ...
- 9,132
40
votes
1
answer
2k
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Curious $q$-analogues
Consider the Fibonacci polynomials
$$F_n (x) = \sum_{j = 0}^{\left\lfloor {n/2} \right\rfloor }\binom{n-j}{j} x^{n - 2j} $$
and the Lucas polynomials
$$L_n (x) = \sum_{j = 0}^{\left\lfloor {n/2} \...
- 5,622
22
votes
4
answers
3k
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Are the q-Catalan numbers q-holonomic?
The generating function $f(z)$ of the Catalan numbers which is characterized by $f(z)=1+zf(z)^2$ is D-finite, or holonomic, i.e. it satisfies a linear differential equation with polynomial ...
- 5,622