Questions tagged [q-analogs]
The q-analogs tag has no usage guidance.
101 questions
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 ...
- 18.4k
7
votes
1
answer
320
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 $n\geq 0$:
$$
(-1)^{...
- 1,430
7
votes
1
answer
253
views
Enumerating subspaces of $\mathbb{F}_q^n$ in terms of words and inversions
When $q$ is a prime power, then on the one hand the $q$-binomial coefficient $\binom{n}{k}_q$ equals the number of $k$-dimensional subspaces of $\mathbb{F}_q^n$, and on the other hand it is the ...
- 63.1k
7
votes
1
answer
325
views
Looking for a $q$-analogue of a binomial identity
The following identity is well-known and there are a few proofs to it (see Bijective proof problems, by R P Stanley, for this and similar formulae):
$$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}=4^n \...
- 43.2k
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 $$D(k,n,m)= \det\left(c(k,...
- 5,622
7
votes
0
answers
280
views
A recursion which defines polynomials with integer coefficients?
Let $[n]=1+q+\dots+q^{n-1}$ and $u(n)=\prod_{j=1}^n \gcd([j],[n])$.
Define
$$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]}$$ with $...
- 5,622
7
votes
0
answers
235
views
$q$-crystals - is there such a thing?
There are several important facts that I first heard about here on MO. One of the most enlightening of these is that $\mathscr D$-modules on a scheme $X$ may be viewed as sheaves on the groupoid of ...
- 18.4k
6
votes
1
answer
644
views
q-analog of the matrix exponential
I am a fan of the Matrix exponential $\exp(X)$, defined for any complex matrix $X$ by
\begin{equation*}
\exp(X) := \sum_{k \ge 0} \frac{X^k}{k!}.
\end{equation*}
I have a fleeting acquaintance with ...
- 28.6k
6
votes
1
answer
340
views
Inequality for functions on [0,1], continued
Let $0<a<1,\; \psi_a(x)=\displaystyle \prod_{j=0}^\infty (1-a^jx).$ For each $ k\in \mathbb{N},$ set
$$f_k(a;x):=\frac{x^k}{(1-a)(1-a^2)\dots (1-a^k)}\,\psi_a(x).$$
Question. Is it true that, ...
- 783
6
votes
1
answer
689
views
Q-binomials at roots of unity
As the title says, given a general $q$-binomial $\binom{n}{k}_q$,
is there some general result regarding its value at a root of unity, $q = \exp(2\pi i r/N)$?
- 15.8k
6
votes
0
answers
112
views
Bijection between forests and skew SYT + Cyclic sieving
Consider the two-row skew shape $\lambda_n = (2n+1,n)/(1)$.
The number of standard Young tableaux of this shape is
$\binom{3n}{n}-\binom{3n}{n-2}$ (since one can easily biject this to the set of non-...
- 15.8k
6
votes
0
answers
227
views
Gaussian coefficients identity
I am having difficulty showing the equivalence between (11) and (15) of Delsarte - Association schemes and $t$-designs in regular semilattices. It is somehow an application of Möbius inversion, but I ...
6
votes
0
answers
214
views
Looking for a combinatorial proof for an identity involving $q$-Catalan triangles
Let $C_n=\frac1{n+1}\binom{2n}n$ be the Catalan numbers. Following my earlier post on MO, one fine colleague asked me if there is a $q$-analogue of the identity formed by the so-called Shapiro's ...
- 43.2k
6
votes
0
answers
132
views
Q-analogue of an inequality
Pick integers $b\geq a \geq 0$ and $k\geq j\geq 0$.
It is not super-difficult to prove the inequality
$$
\binom{kb}{ka}^j \geq \binom{jb}{ja}^k.
$$
This is actually quite a nice inequality that was ...
- 15.8k
6
votes
0
answers
203
views
Conjecture for a certain Cauchy-type determinant
Given the Cauchy-like matrix
$$
\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}}
{\...
- 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,...
- 5,622
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 ...
5
votes
1
answer
2k
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 ...
- 53
5
votes
2
answers
635
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
$$\lim_{q\to1}\frac{d(n)}{(q-1)^\binom{n+...
- 5,622
5
votes
1
answer
401
views
$q$-analog of an integral from quantum field theory?
This question has been completely reformulated and a new property for the function $f_q$ has been added due to a series of helpful comments by fedja.
Consider the integral from quantum field theory ...
- 5,624
5
votes
1
answer
178
views
A $q$-analogue of a characterization of polynomials by binomial coefficients
Considering the binomial coefficient $\binom{x}{m}$ as a polynomial in $x$, the span of $\binom{x}{0}, \binom{x}{1}, \ldots, \binom{x}{d}$ is exactly the polynomials of degree $\le d$. A closely ...
- 11.2k
5
votes
0
answers
388
views
Is a basic hypergeometric function ${}_2\phi_1(a, b; c; q, z)$ a meromorphic function in $z$?
Here a basic hypergeometric function is the analytic continuation of the basic hypergeometric series (or called the $q$-hypergeometric series)
$$
{}_2\phi_1(a, b; c; q, z) = \sum^{\infty}_{n = 0} \...
- 133
4
votes
1
answer
293
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 ...
- 533
4
votes
1
answer
388
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 $$\prod_{i=0}^{A-1} \prod_{j=0}^{B-1} \frac{1}{1-t q^{i+j}} = \...
- 952
4
votes
1
answer
168
views
Discriminants of some $q$-analogs of $(1+x)^n$
Let $[n]_q=1+q+\dots +q^{n-1}$, $ {[n]_q}! =[1]_q [2]_q \dots [n]_q$ and $\binom{n}{j}_q = \frac{[n]_q!}{[j]_q![n-j]_q!}$ be the usual $q$-notation.
Consider the polynomials $p_n(q,r,x)= \sum_{j=0}^n ...
- 5,622
4
votes
2
answers
292
views
How to prove that $\sum_{i=0}^n\frac{(a;q)_i}{(q;q)_i}\frac{(b;q)_{n-i}}{(q;q)_{n-i}}a^{n-i}=\frac{(ab;q)_n}{(q;q)_n}$?
By Cauchy identity, $${}_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,$$
we can obtain the $q-$analogue of $(1-z)^{-a}(1-z)^{-b}=...
- 139
4
votes
1
answer
672
views
Are the following q-Genocchi numbers known?
The sequence of Genocchi numbers
${({G_{2n}})_{n \ge 0}}=$ $(0,1,1,3,17,155,2073,...)$
can be defined by the generating function
$z\frac{{1 - {e^z}}}{{1 + {e^z}}} = \sum {{{( - 1)}^n}{G_{2n}}\frac{...
- 5,622
4
votes
0
answers
108
views
Quantum version of Kostant's basis of ℤ-form of U(𝔤)
Kostant showed that the subring of $\mathcal U(\mathfrak{sl}_2)$ generated by the divided powers $e^c/c!$ and $f^a/a!$ has a $\mathbb Z$-basis given by the elements $\frac{f^a}{a!}\binom hb \frac{e^c}{...
- 71
4
votes
0
answers
113
views
Positivity of q-analogs of central binomial coefficients?
With the usual $q-$notations
$[n]_q=1+q+\cdots+q^{n-1}=\frac{\,\,1-q^n}{1-q},$
$[n]_q!=[1]_q[2]_q\cdots[n]_q$ and
$\binom{n}k_q=\frac{[n]_q!}{[k]_q!\cdot[n-k]_q!}$
let
$$b(n,k,r,q)=\det\left(q^{r\...
- 5,622
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}...
- 5,622
3
votes
2
answers
187
views
Algorithms (or packages) to find recurrence relations for given sequence of q-polynomials?
Assume we have sequence of polynomials : $P_i(q)$ - each term is polynomial in $q$. (With integer coefficients, but hopefully it is not important).
We expect that there exists recurrence relation a ...
- 24.9k
3
votes
1
answer
253
views
What is the value of this sum involving q-binomials?
Let $n\ge 2r$ be positive integers.
Is there a closed form for following finite summation involving in q-binomial coefficients
$$\sum_{s=0}^r(-1)^sq^{\frac{s(s+1)}{2}}{n-2r+s\brack n-2r}_q{n\brack r-...
- 1,093
3
votes
1
answer
186
views
Is there a $q$-analogue to Shapiro's convolution identity?
Let $C_n=\frac1{n+1}\binom{2n}n$ denote the Catalan numbers.
This question is motivated by the (unanswered) MO post by Alexander Burstein and my own (answered by Fedor Petrov) MO post.
Specifically, ...
- 43.2k
3
votes
1
answer
348
views
Closed form for a simple hypergeometric $q$ series
I've run across an interesting hypergeometric $q$-series that I feel must have been found before:
$\sum_{n=0}^{\infty}(-1)^n$$\frac{e^{n b y}}{\prod_{k=1}^{n}(e^{\pi k b^2}-e^{\pi k b^{-2}})} = \...
- 151
3
votes
1
answer
1k
views
Does $\sum_{n=-\infty}^\infty (bq^n,p/aq^n;p)_\infty z^n q^{n(n-1)/2}$ have a closed form?
The formula
$$
\small\sum_{n=-\infty}^\infty (bq^n,p/aq^n;p)_\infty
z^n q^{n(n-1)/2}=\frac{(-z,-q/z;q)_\infty}{\ln\frac{1}{q}}\int\limits_0^\infty\frac{\left(bt/z,pz/at;p\right)_\infty}{\left(-t,-...
- 5,624
3
votes
0
answers
80
views
Applications of q-Lagrange inversion
I was reading a text on q,t-Catalan numbers and Diagonal Harmonics by Haglund, where they mention the following $q$-analogue of Lagrange Inversion, taken from Page 53:
Let $e_n, h_n$ denote the ...
- 31
3
votes
0
answers
136
views
A recursion involving binomial coefficients: looking for a q-analog
Let $a_n := \frac{1}{2n+1}\binom{3n}{n}$.
Then it is known that (one can find references in the OEIS for this.)
$$
a_n = \sum_{\substack{i,j,k \geq 0 \\ i+j+k=n-1} } a_i a_j a_k.
$$
Is there a natural ...
- 15.8k
3
votes
0
answers
106
views
Does the Riemann characterization of the hypergeometric function have a q-analog?
This question is inspired by another recent one here, Characterization of the hypergeometric function. The latter is about the classical result of Riemann characterizing the hypergeometric functions ...
- 18.4k
2
votes
2
answers
236
views
$q$-factorial coefficient asymptotics
Consider the $[n]!_q = \prod\limits_{k = 1}^{n} \frac{q^k - 1}{q - 1} = \sum\limits_{k = 0}^{\binom n 2} c_k q^k$ and let $\{f_n\}_{n \in \mathbb{N}}$ be the sequence of the functions on $[0; 1]$ ...
- 123
2
votes
1
answer
109
views
Does this q-analogue have a nice closed form? [closed]
Let $[n]_q=1+q+\cdots+q^{n-1}$. Is there a nice closed form of $\sum_{s=1}^i[s]_{q}$? One would expect that the answer will be some q-analog of $\frac{i(i+1)}{2}$, since $\sum_{s=1}^i s=\frac{i(i+1)...
- 470
2
votes
1
answer
257
views
Major index generating polynomial for border-strip tableaux
The Question in its original form has been answered, but there is a follow-up, see the end of the post.
A border-strip is a skew Young diagram that does not contain a $2 \times 2$-box. A border-strip ...
- 95
2
votes
2
answers
356
views
What partial sum formulae exist for this basic hypergeometric series?
I've run into:
$$\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$$
I am interested mostly in the cases where $a = 1$ or $ a = 2$
Things I'...
- 131
1
vote
1
answer
191
views
Special values of continuous q - Hermite polynomials
The continuous $q-$Hermite polynomials are defined by
$${H_{n + 1}}(x|q) = 2x{H_n}(x|q) +( {q^n}-1){H_{n - 1}}(x|q)$$
with initial values ${H_{ - 1}}(x|q) = 0$ and ${H_0}(x|q) = 1.$
Cf. e.g. http://...
- 5,622
1
vote
1
answer
232
views
Looking for q-analog of derangement anagrams for a word
I have already known QPermutationDerangement:
It describes the distribution
$$
d_n(q)=\sum_{\sigma \in D_n} q^{\operatorname{maj}(\sigma)}
$$
Where we sum over all derangements of an $n$ element set.
...
- 175
1
vote
0
answers
156
views
Is anything known about the derivative of the quantum dilogarithm?
Faddeev's noncompact quantum dilogarithm is the function defined by
$$
\Phi_{\mathsf b}(z) =
\exp
\int_{\mathbb{R} + i\varepsilon}
\frac{
e^{-2i zw}
}{
4 \sinh(w \mathsf b ) \sinh(w/\...
- 2,533
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 ...
- 11
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{...
- 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 ...
- 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 $...
- 443
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 ...
- 34.3k