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### 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 ...
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### 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 ...
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$\DeclareMathOperator\dim{dim}$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 $K_{\lambda,... 8 votes 7 answers 639 views ### Important combinatorial and algebraic interpretations of the coefficients in the polynomial$[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 501 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. ... 22 votes 2 answers 707 views ### A q-rious identity Let$[x]_q=\frac{1-q^x}{1-q}$,$[n]_q!=_q_q\cdots[n]_q$and${\binom{x}{n}}_{q}=\frac{[x]_q[x-1]_q\cdots[x-n+1]_q }{[n]_q!}$. Computer experiments suggest that $$\det \left(q^\binom{i-j}{2}\... 6 votes 1 answer 312 views ### Lusztig's q-analog of weight multiplicity with product formula For partitions \lambda, \mu \vdash n, the Kostka-Foulkes polynomial K_{\lambda,\mu}(q), a q-analog of the Kostka coefficient K_{\lambda,\mu}, has a combinatorial description, due to Lascoux ... 8 votes 1 answer 219 views ### Prominent examples of q-analogs without known cyclic sieving The cyclic sieving phenomenon is nicely summarized in the following AMS Notices "What is...?" article: https://www.ams.org/notices/201402/rnoti-p169.pdf. In that article, Reiner, Stanton, and White ... 8 votes 1 answer 264 views ### Product of q-analogues Background Recall that the q-analogue [n]_q\in\mathbb Z[q] of a natural number n\in\mathbb N is defined as$$ [n]_q := \frac{q^n -1}{q-1}$$the idea being that formulas involving q will ... 6 votes 0 answers 115 views ### Q-analogue of an inequality Pick integers b\geq a \geq 0 and k\geq j\geq 0. It is not super-difficult to prove the inequality$$ \binom{kb}{ka}^j \geq \binom{jb}{ja}^k. $$This is actually quite a nice inequality that was ... 4 votes 0 answers 110 views ### Positivity of q-analogs of central binomial coefficients? With the usual q-notations [n]_q=1+q+\cdots+q^{n-1}=\frac{\,\,1-q^n}{1-q}, [n]_q!=_q_q\cdots[n]_q and \binom{n}k_q=\frac{[n]_q!}{[k]_q!\cdot[n-k]_q!} let$$b(n,k,r,q)=\det\left(q^{r\... 2 votes 2 answers 207 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 159 views ### Conjecture for a certain Cauchy-type determinant Given the Cauchy-like matrix $$\mathbf X_M(q) = \left[ \frac{2}{\pi} \frac{ \Gamma\!\left(m - \frac{1}{2} \right)\Gamma\!\left(n + \frac{1}{2} \right) }{ \Gamma(m)\,\Gamma(n) } \frac{m-\frac{3}{4}} {\... 2 votes 1 answer 204 views ### Major index generating polynomial for border-strip tableaux The Question in its original form has been answered, but there is a follow-up, see the end of the post. A border-strip is a skew Young diagram that does not contain a 2 \times 2-box. A border-strip ... 6 votes 0 answers 183 views ### A curious q-identity Let [x]_{q}=\frac{1-q^x}{1-q} and \binom{x}{n}_{q} denote a q-binomial coefficient. Let A_n(x,q) be the n\times n matrix with entries$$q^\binom{i-j}{2}\binom{i+j+x}{i-j+1}_{q},$$0 \le i,... 15 votes 0 answers 232 views ### Irreducibility of q-factorial plus 1 Let q be a formal variable and for every positive integer n let$$[n]_q! = 1 (1 + q)(1 + q + q^2) \dotsm (1 + q + \dotsb + q^{n-1})$$be the q-factorial. Is it true that [n]_q! + 1 is an ... 8 votes 0 answers 195 views ### q-analog of (d/dx) \binom{x}{k}? It is not hard to find easy formulas for the derivative of the function \binom{x}{k}, for instance it is not too hard to see (for k an integer) that \frac{d}{dx} \binom{x}{k} = \sum_{i=1}^k \... 2 votes 2 answers 288 views ### What partial sum formulae exist for this basic hypergeometric series? I've run into:$$\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$$I am interested mostly in the cases where a = 1 or a = 2 Things I'... 13 votes 2 answers 528 views ### q as a prime power and a root of unity The number of points on the (n-1)-dimensional projective space P^{n-1}(\mathbb{F}_q) over a finite field \mathbb{F}_q is the q-integer$$[n]_q := \frac{q^n-1}{q-1}.$$In analogy, the number of ... 15 votes 1 answer 591 views ### Schur-Weyl duality and q-symmetric functions Disclaimer: I'm far from an expert on any of the topics of this question. I apologize in advance for any horrible mistakes and/or inaccuracies I have made and I hope that the spirit of the question ... 11 votes 4 answers 715 views ### A divisibility of q-binomial coefficients combinatorially Let a and b be coprime positive integers. Then the number a+b divides the binomial coefficient {a+b \choose a}. I know how to prove this combinatorially - for example after choosing an ordered set ... 9 votes 0 answers 173 views ### For q-analogues of a known curious identity In 2002 I published the folllowing curious combinatorial identity:$$(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.$$My original proof is ... 14 votes 1 answer 633 views ### Is there a lift of the q-Vandermonde identity to some geometric (motivic) identity for Grassmannians over F_q? The q-Vandermonde identity reads:$$ \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} $$... 30 votes 1 answer 1k views ### Mysterious symmetry - in search for a bijection I have a mysterious symmetry that I have not managed to prove. First some definitions (see picture below) Fix a partition that fit in a staircase shape with n rows. There are Catalan(n) such ... 5 votes 1 answer 348 views ### q-analog of an integral from quantum field theory? This question has been completely reformulated and a new property for the function f_q has been added due to a series of helpful comments by fedja. Consider the integral from quantum field theory ... 6 votes 1 answer 303 views ### Inequality for functions on [0,1], continued Let 0<a<1,\; \psi_a(x)=\displaystyle \prod_{j=0}^\infty (1-a^jx). For each k\in \mathbb{N}, set$$f_k(a;x):=\frac{x^k}{(1-a)(1-a^2)\dots (1-a^k)}\,\psi_a(x).$$Question. Is it true that, ... 3 votes 0 answers 91 views ### Does the Riemann characterization of the hypergeometric function have a q-analog? This question is inspired by another recent one here, Characterization of the hypergeometric function. The latter is about the classical result of Riemann characterizing the hypergeometric functions ... 6 votes 1 answer 496 views ### Q-binomials at roots of unity As the title says, given a general q-binomial \binom{n}{k}_q, is there some general result regarding its value at a root of unity, q = \exp(2\pi i r/N)? 5 votes 2 answers 571 views ### Some curious Hankel determinants Let f(n,q)=\prod_{j=1}^na(q^j) for a polynomial a(q) and let d(n)=\det(f(i+j,q))_{i,j=0}^n be its Hankel determinant. Computer experiments suggest that$$\lim_{q\to1}\frac{d(n)}{(q-1)^\binom{n+... 12 votes 1 answer 215 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 106 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)...
As usual, denote $[n]_q=1+q+\cdots+q^{n-1}=\frac{\,\,1-q^n}{1-q}$ and $[n]_q!=_q_q\cdots[n]_q$. Furthermore, we write $$\binom{n}k_q=\frac{[n]_q!}{[k]_q!\cdot[n-k]_q!}.$$ As a follow up on this ...