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Applications of q-Lagrange inversion

I was reading a text on q,t-Catalan numbers and Diagonal Harmonics by Haglund, where they mention the following $q$-analogue of Lagrange Inversion, taken from Page 53: Let $e_n, h_n$ denote the ...
yeetcode's user avatar
1 vote
0 answers
140 views

Is anything known about the derivative of the quantum dilogarithm?

Faddeev's noncompact quantum dilogarithm is the function defined by $$ \Phi_{\mathsf b}(z) = \exp \int_{\mathbb{R} + i\varepsilon} \frac{ e^{-2i zw} }{ 4 \sinh(w \mathsf b ) \sinh(w/\...
Calvin McPhail-Snyder's user avatar
4 votes
1 answer
278 views

Double q-analog of Pochhammer

Has the function $$(z;q_1,q_2)_\infty := \prod_{n_1,n_2=0}^\infty (1-z \, q_1^{n_1} q_2^{n_2}), \quad |q_1|,|q_2|<1$$ been studied in the math literature? For example, does it obey any difference ...
jj_p's user avatar
  • 533
1 vote
0 answers
132 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 ...
Max Alekseyev's user avatar
1 vote
0 answers
51 views

Can I apply $q$-Lagrange Inversion formula?

Now I have equation $F(x) = x \sum_{k\ge 0} g_k F(x) F(qx) \cdots F(q^{k-1} x)$, I need to get the coefficient of $x^n$ in $F(x)$, can I apply $q$-Lagrange Inversion formula to this? Moreover, I have ...
alpha1022's user avatar
7 votes
0 answers
248 views

Hankel determinants for some convolutions of Catalan numbers

Let $c(x)=\frac{1-\sqrt{1-4x}}{2x}$ be the generating function of the Catalan numbers and let $$x^k c(x)^{2k}=(c(x)-1)^k =\sum_{n\geq0}c(k,n)x^n.$$ Consider the determinants $$D(k,n,m)= \det\left(c(k,...
Johann Cigler's user avatar
4 votes
0 answers
106 views

Quantum version of Kostant's basis of ℤ-form of U(𝔤)

Kostant showed that the subring of $\mathcal U(\mathfrak{sl}_2)$ generated by the divided powers $e^c/c!$ and $f^a/a!$ has a $\mathbb Z$-basis given by the elements $\frac{f^a}{a!}\binom hb \frac{e^c}{...
Linus S's user avatar
  • 71
12 votes
0 answers
514 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
6 votes
0 answers
110 views

Bijection between forests and skew SYT + Cyclic sieving

Consider the two-row skew shape $\lambda_n = (2n+1,n)/(1)$. The number of standard Young tableaux of this shape is $\binom{3n}{n}-\binom{3n}{n-2}$ (since one can easily biject this to the set of non-...
Per Alexandersson's user avatar
1 vote
0 answers
141 views

Counting non-zero Gramians of Grassmanians over finite field

In case of $\mathbb{F}_{2}$, we can obtain the number of all reduced row echelon forms (so called Grassmannians) for some m$\times$n full rank matrices by the following gaussian polynomial, $$ \binom{...
mathcat's user avatar
  • 11
7 votes
1 answer
303 views

A curious $q$-series identity on a truncated Euler function

Recall that a $q$-Pochhammer symbol is defined as $$ (x)_n = (x;q)_n := \prod_{l=0}^{n-1}(1-q^l x). $$ I found the following curious $q$-series identity that seems to hold for any $n\geq 0$: $$ (-1)^{...
Henry's user avatar
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10 votes
0 answers
378 views

Has anyone met this "$q$-character" table for $S_4$?

Is anyone aware of the following $q$-character table for the symmetric group $S_4$? \begin{array}{|c|c|c|c|c|c|} \hline \mathrm{conj}\backslash\mathrm{rep} & 2+1+1 & 3+1 & ...
Jeanne Scott's user avatar
  • 1,847
0 votes
0 answers
123 views

Addition formulas for q-analogs of trigonometric functions

Sine and Cosine functions possess notable formulas for addition of angles $$ \sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b) \qquad \text{or} \qquad \cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b). $$ One can ...
Matteo's user avatar
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1 vote
0 answers
87 views

Evaluate $\det[[\lfloor\frac{aj-(a+1)k}n\rfloor]_q]_{1\le j,k\le n}$ and $\det[[\lceil\frac{(a+1)j-ak}n\rceil]_q]_{1\le j,k\le n}$

The $q$-analogue of an integer $m$ is defined by $[m]_q=(1-q^m)/(1-q)$. Note that $\lim_{q\to1}[m]_q=m$. I have formulated the following conjecture on determinants involving the floor function and the ...
Zhi-Wei Sun's user avatar
  • 14.5k
10 votes
2 answers
480 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
3 votes
0 answers
132 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 ...
Per Alexandersson's user avatar
4 votes
1 answer
166 views

Discriminants of some $q$-analogs of $(1+x)^n$

Let $[n]_q=1+q+\dots +q^{n-1}$, $ {[n]_q}! =[1]_q [2]_q \dots [n]_q$ and $\binom{n}{j}_q = \frac{[n]_q!}{[j]_q![n-j]_q!}$ be the usual $q$-notation. Consider the polynomials $p_n(q,r,x)= \sum_{j=0}^n ...
Johann Cigler's user avatar
6 votes
0 answers
223 views

Gaussian coefficients identity

I am having difficulty showing the equivalence between (11) and (15) of Delsarte - Association schemes and $t$-designs in regular semilattices. It is somehow an application of Möbius inversion, but I ...
Leon Bankston's user avatar
9 votes
2 answers
449 views

Is there a nice q-analogue of the Jacobi identity in a quantized enveloping algebra?

In a Lie algebra $\mathfrak{g}$ the Jacobi identity $\newcommand{\bracket}[2]{\left[#1\,#2\right]} \bracket{x}{\bracket{y}{z}} + \bracket{z}{\bracket{x}{y}} + \bracket{y}{\bracket{z}{x}} = 0$ holds. ...
Mike Pierce's user avatar
  • 1,149
3 votes
1 answer
174 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
7 votes
1 answer
311 views

Looking for a $q$-analogue of a binomial identity

The following identity is well-known and there are a few proofs to it (see Bijective proof problems, by R P Stanley, for this and similar formulae): $$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}=4^n \...
T. Amdeberhan's user avatar
6 votes
1 answer
242 views

Enumerating subspaces of $\mathbb{F}_q^n$ in terms of words and inversions

When $q$ is a prime power, then on the one hand the $q$-binomial coefficient $\binom{n}{k}_q$ equals the number of $k$-dimensional subspaces of $\mathbb{F}_q^n$, and on the other hand it is the ...
Martin Brandenburg's user avatar
5 votes
1 answer
169 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 ...
Mark Wildon's user avatar
  • 10.8k
19 votes
1 answer
662 views

What is the groupoid cardinality of the category of vector spaces over a finite field?

For any groupoid, it's groupoid cardinality is the sum of the reciprocals of the automorphism groups over the isomorphism classes. Let us consider the category of vector spaces over a finite field $\...
Asvin's user avatar
  • 7,646
17 votes
1 answer
860 views

Proof of certain $q$-identity for $q$-Catalan numbers

Let us use the standard notation for $q$-integers, $q$-binomials, and the $q$-analog $$ \operatorname{Cat}_q(n) := \frac{1}{[n+1]_q} \left[\matrix{2n \\ n}\right]_q. $$ I want to prove that for all ...
Per Alexandersson's user avatar
6 votes
0 answers
265 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 $...
Johann Cigler's user avatar
8 votes
2 answers
569 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,...
Sam Hopkins's user avatar
  • 22.9k
9 votes
7 answers
724 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
2 answers
558 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. ...
Christian Gaetz's user avatar
22 votes
2 answers
736 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}\...
Johann Cigler's user avatar
8 votes
2 answers
475 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 ...
Sam Hopkins's user avatar
  • 22.9k
8 votes
1 answer
226 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
284 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 ...
Yuri Sulyma's user avatar
  • 1,513
6 votes
0 answers
128 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 ...
Per Alexandersson's user avatar
4 votes
0 answers
113 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\...
Johann Cigler's user avatar
2 votes
2 answers
225 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 $[0; 1]$ ...
DG_'s user avatar
  • 123
6 votes
0 answers
198 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}} {\...
Fred Hucht's user avatar
  • 2,705
2 votes
1 answer
249 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 ...
Joakim Uhlin's user avatar
6 votes
0 answers
190 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,...
Johann Cigler's user avatar
15 votes
0 answers
257 views

Irreducibility of q-factorial plus 1

Let $q$ be a formal variable and for every positive integer $n$ let $$[n]_q! = 1 (1 + q)(1 + q + q^2) \dotsm (1 + q + \dotsb + q^{n-1})$$ be the $q$-factorial. Is it true that $[n]_q! + 1$ is an ...
Penchez's user avatar
  • 341
8 votes
0 answers
241 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 \...
Ratio Bound's user avatar
2 votes
2 answers
336 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'...
user3108815's user avatar
13 votes
2 answers
620 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 ...
Henry's user avatar
  • 1,410
15 votes
1 answer
706 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 ...
Saal Hardali's user avatar
  • 7,549
12 votes
5 answers
814 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 ...
Peter McNamara's user avatar
9 votes
0 answers
191 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 ...
Zhi-Wei Sun's user avatar
  • 14.5k
14 votes
1 answer
758 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: $$ \binom{ a }{ b}_{\!\!q} $$ ...
Alexander Chervov's user avatar
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 ...
Per Alexandersson's user avatar
5 votes
1 answer
392 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 ...
Martin Nicholson's user avatar
6 votes
1 answer
329 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, ...
Deepti's user avatar
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