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15 votes
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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
13 votes
0 answers
349 views

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\...
Johann Cigler's user avatar
12 votes
0 answers
631 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
12 votes
0 answers
502 views

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 ...
Richard Stanley's user avatar
10 votes
0 answers
389 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
  • 2,137
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 ...
Zhi-Wei Sun's user avatar
  • 15.6k
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 \...
Ratio Bound's user avatar
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 ...
მამუკა ჯიბლაძე's user avatar
7 votes
0 answers
252 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
7 votes
0 answers
280 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
7 votes
0 answers
235 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 ...
მამუკა ჯიბლაძე's user avatar
6 votes
0 answers
112 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
6 votes
0 answers
227 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
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 ...
T. Amdeberhan's user avatar
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 ...
Per Alexandersson's user avatar
6 votes
0 answers
203 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
  • 3,756
6 votes
0 answers
191 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
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 ...
Andrius Kulikauskas's user avatar
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} \...
Dong Wang's user avatar
  • 133
4 votes
0 answers
108 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
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
4 votes
0 answers
126 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}...
Johann Cigler's user avatar
3 votes
0 answers
80 views

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
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 ...
Per Alexandersson's user avatar
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 ...
მამუკა ჯიბლაძე's user avatar
1 vote
1 answer
232 views

Looking for q-analog of derangement anagrams for a word

I have already known QPermutationDerangement: It describes the distribution $$ d_n(q)=\sum_{\sigma \in D_n} q^{\operatorname{maj}(\sigma)} $$ Where we sum over all derangements of an $n$ element set. ...
138 Aspen's user avatar
  • 175
1 vote
0 answers
156 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
1 vote
0 answers
55 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
1 vote
0 answers
150 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
1 vote
0 answers
88 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
  • 15.6k
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 $...
Ken Gonzales's user avatar
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 ...
Max Alekseyev's user avatar
0 votes
0 answers
137 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
  • 116