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.
Best Answer
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.