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40 votes
1 answer
2k 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} \...
Johann Cigler's user avatar
22 votes
4 answers
3k 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 ...
Johann Cigler's user avatar
17 votes
1 answer
886 views

Proof of certain $q$-identity for $q$-Catalan numbers

Let us use the standard notation for $q$-integers, $q$-binomials, and the $q$-analog $$ \operatorname{Cat}_q(n) := \frac{1}{[n+1]_q} \left[\matrix{2n \\ n}\right]_q. $$ I want to prove that for all ...
Per Alexandersson's user avatar
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)^{...
Henry's user avatar
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