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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.

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  • $\begingroup$ 1. In the definition of $F_g$, do you mean summation in $d$? Otherwise the definition is not clear: if you sum in $g$, the LHS will depend on $d$ rather than $g$. 2. Where is the Riccati equation in your question? All equations that you wrote are linear. $\endgroup$ Oct 2, 2021 at 14:36
  • $\begingroup$ Yes, summation over d, you are right I only need linear ode. $\endgroup$
    – GGT
    Oct 3, 2021 at 0:50

1 Answer 1

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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

Abramov, S. A., Rational solutions of linear differential and difference equations with polynomial coefficients, U.S.S.R. Comput. Math. Math. Phys. 29, No. 6, 7-12 (1989). ZBL0719.65063.

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.

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  • $\begingroup$ @GGT I'm sorry, I don't really understand your comment or question (I can only guess). Maybe it would help you to write shorter sentences. Here ansatz = parametrized guess for the solution. All information about poles is obtained through the Frobenius method. $\endgroup$ Oct 6, 2021 at 14:03
  • $\begingroup$ I am solving for the family $F_g (x)$ such that $g\geq 0$. Having solved for $F_0 (x)$, I noticed that a change of coordinate gives the rational solution and the leading coefficient of the LHS of the eq (**) is (1-2z) so the rational solution have a denominator (2z-1) and the same is true for all $F_g$ in z coordinate. I will take a look ar Frobenius method. $\endgroup$
    – GGT
    Oct 6, 2021 at 14:43
  • $\begingroup$ Thanks a lot for the answer and direction. $\endgroup$
    – GGT
    Oct 6, 2021 at 14:46

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