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.

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