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 coefficients. 
The generating function $F(z)$ of the $q$-Catalan numbers is analogously characterized by the functional equation $F(z)=1+z F(z) F(qz)$. I suspect that $F(z)$ is not $q$-holonomic, i.e. does not satisfy a linear $q$-differential equation with polynomial coefficients. But I have no proof. Is there a proof in the literature or references which may lead to a proof?


Since there were some misunderstandings I want to clarify the situation.
A power series $F(z)$ is called $q - $holonomic if there exist polynomials $p\_i (z)$ such that $\sum\limits\_{i = 0}^r {p\_i (z)D\_q^i } F(z) = 0$ where $D\_q $ denotes the $q - $differentiation operator defined by $D\_q F(z) = \frac{{F(z) - F(qz)}}{{z - qz}}.$ Equivalently if there exist (other) polynomials such that $\sum\limits\_{i = 0}^r {p\_i (z)F(q^i } z) = 0.$


Let $f(z)$ be the generating function of the Catalan numbers $\frac{1}{{n + 1}}{2n\choose n}$ . Then $f(z) = 1 + zf(z)^2 $ or equivalently $f(z) = \frac{{1 - \sqrt {1 - 4z} }}{{2z}}.$
There are 3 simple $q - $analogues of the Catalan numbers: 
a) The polynomials $C\_n (q)$ introduced by Carlitz with generating function $F(z) = 1 + zF(z)F(qz)$. My question is about these polynomials. Their generating function satisfies a simple equation, but there is no known formula for the polynomials themselves.
b) The polynomials $\frac{1}{{[n + 1]}}{2n\brack n}$ . They have a simple formula but no simple formula for their generating function.
c) The $q - $Catalan numbers $c\_n (q)$ introduced by George Andrews. Their generating function $A(z)$ is a $q - $analogue of $ \frac{{1 - \sqrt {1 - 4z} }}{{2z}}.$ Let $h(z)$ be the $q - $analogue of $sqrt {(1 + z)}$ defined by $h(z)h(qz)=1+z$. Then $A(z)= \frac{1+q}{{4qz}}(1-h(-4qz))$. They have both simple formulas and a simple formula for the generating function. But they are not polynomials in $q.$ 
Both b) and c) are $q$-holonomic. My question is a proof that a) is not $q$-holonomic.



Are the q-Catalan numbers q-holonomic?
The problem is to show that the function $F(z)=1+z+(1+q)z^2+O(z^3)$
satisfying the functional equation $F(z)=1+zF(z)F(qz)$ does not satisfy
$\sum\_{j=0}^{n-1}P\_j(z)F(q^jz)+Q(z)=0$ identically in $z$ for some $n$;
here $P\_j$ and $Q$ are polynomials in both $z$ and $q$.
(Although the original question assumes the homogeneous equation,
$Q(z)=0$, the limiting case $q\to1$ suggests to consider $Q(z)$ more
generally.) In what follows we show that such a functional equation
implies the algebraicity of $F(z)$; this is known to be false.


First of all, switch to the function $G(z)=zF(z)$ which satisfies
$$
G(qz)G(z)=q(G(z)-z).
$$
The problem is then to show that the newer function does not satisfy
$\sum\_{j=0}^{n-1}\tilde P\_j(z)G(q^jz)+\tilde Q(z)=0$ for some $n$.
By applying $z\mapsto q^{-k}z$ we can assume that $\tilde P\_0(z)\ne0$
in this relation. The substitution $z\mapsto qz$ results in the relation
$\sum\_{j=1}^n\hat P\_j(z)G(q^jz)+\hat Q(z)=0$ where $\hat P\_1(z)\ne0$.


The next step is to show, by iterating the functional equation for $G(z)$, that
$$
G(q^nz)\dots G(qz)G(z)=X\_n(z)G(z)-Y\_n(z).
$$
Indeed, we have $X\_0=1$, $Y\_0=0$, and
$$
X\_n(z)G(z)-Y\_n(z)=\bigl(X\_{n-1}(qz)G(qz)-Y\_{n-1}(qz)\bigr)G(z)
$$
implying
$$
X\_n(z)=qX\_{n-1}(qz)-Y\_{n-1}(qz), \quad Y\_n(z)=qzX\_{n-1}(qz)
\qquad\text{for}\quad n\ge1.
$$
By means of the formula we see that $\deg Y\_n$ does not decrease with $n$,
so that $Y\_n\ne0$ for $n\ge1$. Note that our computation implies
$$
G(q^nz)=\frac{X\_n(z)G(z)-Y\_n(z)}{X\_{n-1}(z)G(z)-Y\_{n-1}(z)}
\qquad\text{for}\quad n\ge1.
$$


Substitute the above finding into the equation
$\sum\_{j=1}^n\hat P\_j(z)G(q^jz)+\hat Q(z)=0$ with $\hat P\_1(z)\ne0$.
We obtain
$$
\sum\_{j=1}^n\hat P\_j(z)\frac{X\_j(z)G(z)-Y\_j(z)}{X\_{j-1}(z)G(z)-Y\_{j-1}(z)}
+\hat Q(z)=0.
$$
The term corresponding to $j=1$ is equal to
$$
\hat P\_1(z)\frac{X\_1(z)G(z)-Y\_1(z)}{X\_0(z)G(z)-Y\_0(z)}
=\hat P\_1(z)\frac{qG(z)-qz}{G(z)}.
$$
Note that the denominator $G(z)$ in this expression is not
canceled by the other denominators in the former relation
because $X\_{j-1}(z)G(z)-Y\_{j-1}(z)$ is never divisible by $G(z)$
as $Y\_{j-1}(z)\ne0$ for $j\ge2$. In other words, after multiplying
$$
\sum\_{j=1}^n\hat P\_j(z)\frac{X\_j(z)G(z)-Y\_j(z)}{X\_{j-1}(z)G(z)-Y\_{j-1}(z)}
+\hat Q(z)=0
$$
by $G(z)$ we get an algebraic relation
$$
-qz\hat P\_1(z)+G(z)\cdot\text{rational function in }z,q,G(z)
=0
$$
where the rational function does not involve $G(z)$ as a multiple in the denominator.
Since $\hat P\_1(z)\ne0$, this gives us a nontrivial algebraic relation for $G(z)$.
Thus the function $G(z)$, hence $F(z)$ as well, are algebraic.


**Addition.**
I have never seen Carlitz's function $F(z)$ before, so I was pretty sure that
things like its transcendence had been already established. Especially, since
it is not hard (although required some time from me). Any way, this explains
my reasons for not having a reference to this fact.


If we write
$$
F(z)=F\_q(z)=\sum\_{n=0}^\infty a\_n(q)z^n=\sum\_{n=0}^\infty a\_nz^n=1+z+(1+q)z^2+\dots,
$$
the functional equation $F(z)=1+zF(z)F(qz)$ implies
$$
a\_{n+1}=\sum\_{k=0}^nq^ka\_ka\_{n-k} \quad\text{for}\; n\ge0,
\qquad a\_0=1.
$$
The clear induction on $n$ shows that $a\_{n+1}(q)$ is a polynomial from $\mathbb Z[q]$
with leading term $q^{n(n-1)/2}$. In particular, the denominator of $a\_{n+1}(1/2)$ is
exactly $2^{n(n-1)/2}$.


If the function $F\_q(z)$ were algebraic then its specialization $F\_{1/2}(z)$ should be
algebraic. This would imply that for a certain integer $A$ the $z$-expansion of
$F\_{1/2}(Az)$ has *integral* coefficients. But no $A\in\mathbb Z$ with the property
$A^{n+1}/2^{n(n-1)/2}\in\mathbb Z$ for all $n$ could be given. Therefore, $F\_{1/2}(z)$ and
$F\_q(z)$ in general are transcendental.


