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Suppose $F$ is $q-$holonomic. This implies, if I understand correctly that there exists a natural integer $d$ and $d+1$ polynomials $P_0,\dots,P_d\in\mathbb C[q,z]$, not all zero, such that $\sum_{k=0}^d P_k\frac{\partial^k F}{\partial q^k}=0$.

We can assume that not all polynomials $P_i$ are divisible by $z$. We can thus write $P_i=Q_i+z\tilde P_i$ with $Q_0,\dots,Q_d$ in $\mathbb C[q]$ and not all zero. Denoting by $F_n\in\mathbb N[q]$ the coefficient of $z^n$ in $F=\sum_{n=0}^\infty F_n$, an induction on $n$ shows easily that $F_n$ has degree ${n\choose 2}$ as a polynomial in $q$. This implies that given an arbitrary natural integer $A$ there exists $N$ such that a all coefficients of degree $\geq {n\choose 2}-A$ of $\sum_{k=0}^d Q_k\frac{\partial^kF_n}{\partial q^k}\in\mathbb C[q]$ are zero for $n\geq N$.

Since the coefficients of $F_n$ enumerate Dyck paths weighted by a suitable area, the leading coefficients of $F_n$ stabilize to the sequence $1,1,2,3,5,7,\dots$ of partition numbers. This implies that for a given $B$ all coefficients of degree at least ${n\choose 2}-B$ in $Q_e\frac{\partial^eF_n}{\partial q^e}$ are zero for $n$ huge enough and for $e$ the largest integer $\geq 0$ such that $Q_e\not=0$. Since $B$ is arbitrary, this implies $Q_e\prod_{n=1}^\infty\frac{1}{1-q^n}=0$ contradicting the integrality of the ring that arbitrarily many leading terms of formal power series$Q_e\frac{d^e}{dq^e}\left(q^n\prod_{n=1}^\infty\frac{1}{1-q^n}\right)$ are zero if $n$ is large enough. This is absurd.

Remark that the last argument can be simplified: we don't need convergency of the leading degrees of $F_n$. In fact, the above proof shows that a series $G=\sum_{n=0}^\infty G_nz^n$ with $G_n\in\mathbb C[q]$ is never $q-$holonomic if $\limsup_{n\rightarrow\infty}\frac{\deg_q(G_n)}{n}>1$.

1

Suppose $F$ is $q-$holonomic. This implies, if I understand correctly that there exists a natural integer $d$ and $d+1$ polynomials $P_0,\dots,P_d\in\mathbb C[q,z]$, not all zero, such that $\sum_{k=0}^d P_k\frac{\partial^k F}{\partial q^k}=0$.

We can assume that not all polynomials $P_i$ are divisible by $z$. We can thus write $P_i=Q_i+z\tilde P_i$ with $Q_0,\dots,Q_d$ in $\mathbb C[q]$ and not all zero. Denoting by $F_n\in\mathbb N[q]$ the coefficient of $z^n$ in $F=\sum_{n=0}^\infty F_n$, an induction on $n$ shows easily that $F_n$ has degree ${n\choose 2}$ as a polynomial in $q$. This implies that given an arbitrary natural integer $A$ there exists $N$ such that a all coefficients of degree $\geq {n\choose 2}-A$ of $\sum_{k=0}^d Q_k\frac{\partial^kF_n}{\partial q^k}\in\mathbb C[q]$ are zero for $n\geq N$.

Since the coefficients of $F_n$ enumerate Dyck paths weighted by a suitable area, the leading coefficients of $F_n$ stabilize to the sequence $1,1,2,3,5,7,\dots$ of partition numbers. This implies that for a given $B$ all coefficients of degree at least ${n\choose 2}-B$ in $Q_e\frac{\partial^eF_n}{\partial q^e}$ are zero for $n$ huge enough and for $e$ the largest integer $\geq 0$ such that $Q_e\not=0$. Since $B$ is arbitrary, this implies $Q_e\prod_{n=1}^\infty\frac{1}{1-q^n}=0$ contradicting the integrality of the ring of formal power series.

Remark that the last argument can be simplified: we don't need convergency of the leading degrees of $F_n$. In fact, the above proof shows that a series $G=\sum_{n=0}^\infty G_nz^n$ with $G_n\in\mathbb C[q]$ is never $q-$holonomic if $\limsup_{n\rightarrow\infty}\frac{\deg_q(G_n)}{n}>1$.