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
$$F\_{g}(x) := \sum\_{g\geq 1} m\_g (d) x^d$$


$$
\begin{split}
x{\frac {\rm d}{{\rm d}x}}F\_g \left( x \right) &-4\,{x}^{2}{\frac {\rm d}
{{\rm d}x}}F\_g \left( x \right) +2\,xF\_g \left( x \right) \\
&= \left(x{\frac {\rm d}{{\rm d}x}}\right)^3F\_{g-1} \left( x \right) -2\,\left({x}{\frac {\rm d}
{{\rm d}x}}\right)^2F\_{g-1} \left( x \right) +\,xF\_{g-1} \left( x \right) 
\end{split}\label{2}\tag{$\*\*$}
$$
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 $F\_0 = C\sqrt{(1-4x)}$.


Now I can do a change of coordinate
$$x(z) = z-z^2 $$
The solution become rational. That is solution is $(1-2z)$.
From the equation \eqref{3} we can determine that the rational solution can have poles at $z=1/2$.
My question can we determine all other $F\_g$ in coordinates z and it's rational, and determine the poles or $F\_g$ and it's order? How much information we can conclude about $F\_g$ for $g>0$ having the solution of $F\_0 (z)$ and is rational. Any reference will be very helpful.



Rational solution of differential equation
The recurrence $(\*)$ gives the correct form of $(\*\*)$ as $$(1 - 4x) F'\_g(x) + 2 F\_g(x) = x^2 F'''\_{g-1}(x) + x F''\_{g-1}(x) \tag{\*\*}$$


The boundary condition which gives the the correct form of $F\_0$ is $$(1 - 4x) F'\_0(x) + 2 F\_0(x) = 1 \tag{\*\*\*}$$ yielding $F\_0 = \frac{1 + 2C\_0 \sqrt{1 - 4x}}2$ and we further require $2C\_0 = -1$ to give the offset Catalan numbers.


Then we get $$F\_1 = \frac{x^2 + C\_1(1-4x)^3}{(1-4x)^{5/2}}$$ where we desire $C\_1 = 0$; $$F\_2 = \frac{x^2(9x^2 + 20x + 1) + C\_2(1-4x)^6}{(1-4x)^{11/2}}$$ where we desire $C\_2 = 0$; $$F\_3 = \frac{x^2(450x^4 + 3080x^3 + 1770x^2 + 136x + 1) + C\_3(1-4x)^9}{(1-4x)^{17/2}}$$ where we desire $C\_3 = 0$, and I'm sure you can take it from there.


