The following is a generating function in $x,h$ with infinite parameters
$q_1,q_2\ldots,$ and $w_1, w_2,\ldots$.

$$\Psi(x, h)= \sum_{d=0}^{\infty} s_{(d)} (q_1, q_2, \ldots) \exp \bigg( \sum_{r=1}^{\infty} \frac{w_r h^r}{(r+1)!} \left[ \bigg(d-\frac{1}{2}\bigg)^{r+1} - \bigg(-\frac{1}{2}\bigg)^{r+1} \right] \bigg) x^d h^{-d} $$

I want to find the annihilator of the above generating function. Let me list down the cases when it's known So when $q_1 =q_1$ and all other $\{q_i =0\mid i\geq 2\}$ and $w_r=w_r$ and $\{w_i =0\mid i\geq 2\}$ $$\big[ h x\frac{\partial}{\partial x}-q_1\hat{x}\mathcal{A} \big] \Psi(x, h)=0 $$ where $$\mathcal{A}:=x^{\frac32}exp\Big(w_{r}\frac{x^{-1}\sum_{i=0}^{r}(hx\frac{\partial}{\partial x})^{i}x(hx\frac{\partial}{\partial x})^{r-i}}{r+1}\Big)x^{-\frac{1}{2}}$$ $\hat{x} = x$ and $\hat{y} = h x \frac{\partial}{\partial x}$. As $s_{d}(q_1)=\frac{q_1^d}{d!} $ and $\mathcal{A}x^d=exp(\left[ \bigg(d+\frac{1}{2}\bigg)^{r+1} - \bigg(d-\frac{1}{2}\bigg)^{r+1} \right])x^{d+1}$ Hence it annihilates.

Let's fix $q_1,q_2,\ldots,q_e$ and $w_1, \ldots w_{e-1}$ Is this true that

$$ \big[ h x\frac{\partial}{\partial x}-(q_1\hat{x}\mathcal{A}+q_2 \hat{x}^{2}\mathcal{A}^2+\ldots +q_e \hat{x}^{e}\mathcal{A}^e \big] \Psi(x, h) = 0, $$ where $$\mathcal{A}:=x^{\frac32}exp\Big(\sum_{r=1}^{e-1}w_{r}\frac{x^{-1}\sum_{i=0}^{r}(hx\frac{\partial}{\partial x})^{i}x(hx\frac{\partial}{\partial x})^{r-i}}{r+1}\Big)x^{-\frac{1}{2}}$$ $\hat{x} = x$ and $\hat{y} = h x \frac{\partial}{\partial x}$?

I believe there should be an annihilator as each of the series $\sum_d s_d(q_1,\ldots q_r)x^d$ is annihilated by a holonomic function and similarly for $\sum_d \exp \bigg( \sum_{r=1}^{e-1} \frac{w_r h^r}{(r+1)!} \left[ \bigg(d-\frac{1}{2}\bigg)^{r+1} - \bigg(-\frac{1}{2}\bigg)^{r+1} \right]x^d$ Though the annihilator is not holonomic if it was then I could be Hadamard product to conclude the existence of the annihilator.