The q-analogs tag has no wiki summary.

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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
...

**1**

vote

**1**answer

119 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. ...

**2**

votes

**1**answer

389 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 ...

**8**

votes

**1**answer

375 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 - ...

**3**

votes

**1**answer

569 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 {{{( - ...

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votes

**0**answers

242 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 ...

**10**

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**5**answers

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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 ...

**11**

votes

**2**answers

466 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

380 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 ...

**12**

votes

**1**answer

419 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 $n$ ...

**2**

votes

**1**answer

252 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}} = ...

**20**

votes

**6**answers

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 ...

**26**

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**0**answers

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} ...

**18**

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**4**answers

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 ...

**14**

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**2**answers

361 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( ...