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### 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**

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131 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 ...

**11**

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**1**answer

232 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. ...

**11**

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**1**answer

420 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 $n$ ...

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23 views

### How to prove ${}_2\phi_1(q^{-n},b;c;q,q)=\frac{(c/b;q)_n}{(c;q)_n}b^n.$

Firstly, we have already known the one of $q$-analogues of Vandermonde's formula, which is
$${}_2\phi_1(q^{-n},b;c;q,cq^n/b)=\frac{(c/b;q)_n}{(c;q)_n}.$$
And there is a hint, when we change the order ...

**3**

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55 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

296 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]...

**6**

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216 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 ...

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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}$...

**7**

votes

**1**answer

626 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 $W_1 \cap W_2$ ...

**18**

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**1**answer

377 views

### 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 ...

**8**

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154 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\...

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425 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
...

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vote

**1**answer

135 views

### 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://...

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**1**answer

659 views

### 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 ...

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391 views

### 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{\...

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votes

**1**answer

605 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{...

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317 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 $...

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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 ...

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582 views

### 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 ...

**4**

votes

**1**answer

412 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 ...

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**1**answer

488 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 $n$ ...

**3**

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**1**answer

271 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}} = \...

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2k views

### 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 ...

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1k views

### 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 ...

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votes

**1**answer

518 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 ...

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1k views

### 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} \...

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1k views

### 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 ...

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388 views

### Derangements and q-variants

Everybody knows that there are $D_n=n! \left( 1-\frac1{2!}+\frac1{3!}-\cdots+(-1)^{n}\frac1{n!} \right)$ derangements of $\{1,2,\dots,n\}$ and that there are $D_n(q)=(n)_q! \left( 1-\frac{1}{(1)_q!}+\...