7
votes
0answers
116 views

Infinite sum involving $q$-adic representations of whole numbers and $q$-factorial numbers

Let $q \in \mathbb{N}_{\geq 2}$. For $n \in \mathbb{N}_0$, let $$\mathrm{fac}_q(n) := \prod_{i=1}^n (1+q+\dots+q^{i-1}) = \prod_{i=1}^n \frac{q^i-1}{q-1}$$ be the $q$-factorial of $n$. In particular, ...
4
votes
1answer
132 views

Intuition behind the Jacobi triple product

Jacobi's triple product identity states that: $\displaystyle \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}} = \prod_{n = 1}^{\infty}(1 - q^{2n})(1 + zq^{2n - 1})(1 + z^{-1}q^{2n - 1})$ I've seen a messy ...
6
votes
2answers
201 views

Example of something easier to count with $q$-analog?

Are there any known examples of combinatorial objects that become easier to count by considering some kind of $q$-analog? It seems to me that it might be impossible for the problem of computing the ...
6
votes
1answer
233 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 ...
1
vote
1answer
85 views

Combinatorial identity. Using echelon matrices.

Determine the exponents $e_i$ s.t. the following identity is correct. $$\sum\limits_{i=0}^k q^{e_i} {\binom mi}_q {\binom{n}{k-i}}_q = {\binom{n+m}{k}}_q$$ Note: When $q=1$ the equation reduces to a ...
6
votes
3answers
215 views

How to prove it? (one of the Rogers-Ramanujan identities)

Prove the following identity (one of the Rogers-Ramanujan identities) on formal power series by interpreting each side as a generating function for partitions: ...
4
votes
0answers
101 views

A general Combinatorics problem (Coefficients of the q factorial)

I was solving a combinatorics problem when I encountered difficulties. The problem was: $x_1 \in \{0,1\}$ $x_2 \in \{0,1,2\}$ . . $x_{n-1}\in\{0,1,2..,n-1\}$ We have to find the number of ways ...
3
votes
4answers
1k views

Proving q-binomial identities

I was wondering if anyone could show me how to prove q-binomial identities? I do not have a single example in my notes, and I can't seem to find any online. For example, consider: ${a + 1 + b \brack ...
8
votes
0answers
172 views

Different notions of q-numbers

It seems that most of the literature dealing with q-analogs defines q-numbers according to $$[n]_q\equiv \frac{q^n-1}{q-1}.$$ Even Mathematica uses this definition: with the built-in function QGamma ...
4
votes
1answer
212 views

Reciprocity Law of the Gaussian (or $q$-Binomial) Coefficient

It is a standard exercise in combinatorics to show that the binomial coefficient satisfies the reciprocity law $\binom{-n}{k} = (-1)^k \binom{n+k-1}{k}$ for $n, k \geqslant 0$, which is the multiset ...
4
votes
0answers
58 views

Necessary and sufficient condition for $f(q^n)$ to be in $\mathbb{Z}[q,q^{-1}]$ when $f\in\mathbb{Q}(q)[x]$?

In this question, user begins shows that, for each $k\in \mathbb{N}$, there is a unique polynomial $P_k(x)$ of degree $k$ whose coefficients are in $\mathbb{Q}(q)$, the field of rational functions, ...
7
votes
2answers
255 views

q-Analogue of the formula $x^n=\sum_k\left\{n\atop k\right\}(x)_k$.

The stirling numbers of the second kind satisfy the formula $x^n=\sum_k\left\{n\atop k\right\}(x)_k$, where $(x)_k$ is the falling factorial. Consider the $q$-analog recursive definition of the ...
4
votes
2answers
279 views

Recurrence for $q$-analog for the Stirling numbers?

I read in some papers that the Stirling numbers (of the second kind) have a natural $q$-analog $S_q(n,k)$, which satisfy the recurrence $$ S_q(n,k)=(k)_qS_q(n-1,k)+q^{k-1}S_q(n-1,k-1) $$ with the ...
1
vote
1answer
186 views

Inferring Jacobi Triple product from $q$-binomial theorem?

Quite a while ago I asked a question about deducing the Jacobi triple product from the $q$-binomial theorem. In fact, from the $q$-binomial theorem $$ ...
2
votes
1answer
166 views

Is the following product of $q$-binomial coefficients a polynomial in $q$?

$$\frac{\binom{n}{j}_q\binom{n+1}{j}_q \cdots\binom{n+k-1}{j}_q}{\binom{j}{j}_q\binom{j+1}{j}_q\cdots\binom{j+k-1}{j}_q}$$ where $n,j,k$ are non-negative integers.
6
votes
1answer
253 views

Intermediate step in deducing Jacobi's triple product identity.

An intermediate step deduces Jacobi's triple product identity by taking the $q$-binomial theorem $$ \prod_{i=1}^{m-1}(1+xq^i)=\sum_{j=0}^m\binom{m}{j}_q q^{\binom{j}{2}}x^j $$ and deducing $$ ...
1
vote
2answers
338 views

Representing the $q$-binomial coefficient as a polynomial with coefficients in $\mathbb{Q}(q)$?

Trying a bit of combinatorics this winter break, and I don't understand a certain claim. The claim is that for each $k$ there is a unique polynomial $P_k(x)$ of degree $k$ whose coefficients are ...
5
votes
1answer
228 views

Deriving Cauchy's identity from the $q$-binomial theorem?

Cauchy's identity states that $$ \prod_{i\geq 0}\frac{1-axq^i}{1-xq^i}=\sum_{n\geq 0}\frac{(1-a)(1-aq)\cdots(1-aq^{n-1})}{(1-q)(1-q^2)\cdots(1-q^n)}x^n. $$ Is it possible to somehow derive this ...
3
votes
1answer
176 views

What's the reasoning for this recurrence on $q$-multinomial coefficients?

I'm familiar with the recurrence for binomial coefficients based on Pascal's triangle. However, in general, there is the recurrence for $q$-multinomial coefficients given by $$ ...
13
votes
4answers
494 views

Intriguing polynomials coming from a combinatorial physics problem

For real $0<q<1$, integer $n >0 $ and integer $k\ge 0$, define $$[k, n]_q \equiv -\sum_{m=1}^{n} q^{m(k+1)} (q^{-n}; q)_m = -\sum_{m=1}^{n} q^{m(k+1)} \prod_{l=0}^{m-1} (1-q^{l-n})$$ ...