# Rational solution of differential equation

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.

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.