The following is a recursion for one point monotone Hurwitz numbers $$ d \, m_g(d) = 2(2d-3) \, m_g(d-1) + d(d-1)^2 \, m_{g-1}(d)\label{1}\tag{$*$} $$ with initial condition $m_0 (1) =1$ and some of the other numbers are $ m_0 (2) = 1, m_1 (3) =10$. Let denote the generating function by $$F_{g}(x) := \sum_{d\geq 1} m_g (d) x^d$$

$$ \begin{split} x{\frac {\rm d}{{\rm d}x}}F_g \left( x \right) &-4\,{x}^{2}{\frac {\rm d} {{\rm d}x}}F_g \left( x \right) +2\,xF_g \left( x \right) \\ &= \left(x{\frac {\rm d}{{\rm d}x}}\right)^3F_{g-1} \left( x \right) -2\,\left({x}{\frac {\rm d} {{\rm d}x}}\right)^2F_{g-1} \left( x \right) +\,xF_{g-1} \left( x \right) \end{split}\label{2}\tag{$**$} $$ Now we put the condition that $F_g (x) =0 $ for $g<0$ hence using \eqref{2} we get $$ x{\frac {\rm d}{{\rm d}x}}F_0 \left( x \right) -4\,{x}^{2}{\frac {\rm d} {{\rm d}x}}F_0 \left( x \right) +2\,xF_0 \left( x \right)=0\label{3}\tag{$***$} $$ We get $F_0 = C\sqrt{(1-4x)}$. With the change of coordinates $x(z) = z -z^2$ the we get $F_0 (z)$ to be a rational solution hence in coordinate $z$ for this particular equation we get all the solution $F_g (z)$ to be rational. I hope I am not wrong here.

My question is the following given a general one point recursion of type (*) that is say recursion of type $$\sum_{i,j}^{i_{max}, j_{max}} p_{ij} (d)n_{g-i}(d-j) $$ From this one point recursion we can get differential equation of type (**), for the generating function $$F_{g}(x) := \sum_{d\geq 1} n_g (d) x^d$$ So what constraints do we need to put on the polynomials $p_{ij}(d)$ such that the solution will be rational for the differential equations. In the example I gave is the differential equation is Linear ode, so is there any reference regarding rational solutions and their relation to the poles of solution.