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.
Best Answer
I must say that I don't much understand the motivation coming from number theory, but your question about rational solutions of ODEs has a definite answer, provided the equation has polynomial or rational coefficients. A standard reference is
The main idea is straight forward and I will summarize it here. The solution of a homogeneous linear ODE with rational coefficients (which can always be made polynomial by clearing denominators) can have a pole (including the poles at infinity) only at a regular singular point, and only if the Frobenius method reveals integer negative critical exponents. Taking the most negative integer critical exponents at each regular singular point gives you a universal polynomial denominator $Q$, so that any rational solution can be written as $P/Q$, with $P$ polynomial. The Frobenius analysis at infinity also gives you an upper bound on the degree of $P$, which makes $P/Q$ an ansatz for the most general rational solution with finitely many parameters. Now, it is only a matter of plugging in the ansatz into the ODE and expanding in a power series. The preceding analysis converts the equation into a finite dimensional linear algebra problem on the parameters of the $P/Q$ ansatz.