My question is about finding the annihilator of a series. Let me begin with what is known and then ask my question. Let $s_d(\frac{q_1}{h},\ldots )$ denote schur function for partition $\lambda =[d]$ expanded in a power sum basis $q_i$. For the following series
$$F(x):=\sum_{d\geq 0 } s_{(d)}(\frac{q_{1}}{h} , \frac{q_{2}}{h},\ldots,)exp\Big(h^r\frac{(d-1/2)^{r+1}-(-1/2)^{r+1}}{r+1}\Big)x^d $$ with replacing $q_2 , q_3\ldots =0 $ we have $$\Big(h\ x \frac{\partial}{\partial x} - q_1\mathcal{A}\Big)F(x)=0.$$ where $$\mathcal{A}:=x^{\frac32}exp\Big(\frac{x^{-1}\sum_{i=0}^{r}(hx\frac{\partial}{\partial x})^{i}x(hx\frac{\partial}{\partial x})^{r-i}}{r+1}\Big)x^{-\frac{1}{2}}$$ The above claim is not difficult to see as $$\mathcal{A}x^d = exp\Big(h^r\frac{(d+\frac12)^{r+1}-(d-\frac12)^{r+1}}{r+1}\Big)x^{d+1}$$ and $s_d (\frac{q_1}{h})=\frac{q_{1}^{d} }{h^d d!}$ So taking the ratio of two consequitive terms in $F(x)$ give us the annihilator.
Now I want to generalise this procedure for the finite number of $q_i$ being nonzero. $H(x)=\sum_d s_{(d)}(\frac{q_{1}}{h} , \frac{q_{2}}{h},\ldots,)$ is holonomic and also we have an explicit equation of order 1 and degree say r if $q_r$ is the nonzero term and after which all zero. Also, we know that there exists annihilator for $$\sum_d exp\Big(h^r\frac{(d-1/2)^{r+1}-(-1/2)^{r+1}}{r+1}\Big)x^d $$ by using the operator $\mathcal{A}$ hence $\color{red}{\text{We can conclude there exists an annihilator of $F(x)$ by using Hadamard product of two series. }}$ Of course, I have the problem with the above statement as the Annihilator that involve $\mathcal{A}$ is holomorphic hence it will not satisfy the closure properties of the holomorphic function.
I can see that there still exists an Annihilator for $F(x)$ that is following the closure properties of the holomorphic function. So we can generalise the properties of closure properties of the holomorphic function to a more general setting where can include this?
Is there is an explicit annihilator for finitely many or infinitely many nonzero $q_i$.
