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.