Recurrence for a sequence defined by an algebraic generating function

Question

I define the sequence $\{a_n\}_{n \ge 0}$ by the generating function $$ G(g, \xi) = \sum_{n \ge 0} a_n g^n = \frac{1}{g} \, \frac{4\bar g \, (2\bar g^2 - 3\bar g + \xi/2+1/2)} {(\xi-1) \, (2\bar g - \xi)}, $$ where $\bar g$ satisfies the cubic algebraic equation $$ \frac{\bar g \, (2\bar g - \xi/2 -1/2 )^2} {(2\bar g - 1)(2\bar g - \xi)} = g \, . $$

The first few coefficients (if we expand $G(g, \xi)$ in series and looking at the numerator ) are $ 1,\; 1,\; 4,\; 1,\; 1,\; 10,\; 22,\; 10,\; 1,\; 1,\; 18,\; 88,\; 150,\; 88,\; 18,\; 1,\; \dots $

Question I would like to find a linear recurrence (or linear differencial recurrence) satisfied by the coefficients $a_n$, for example of the general form $$ c_0(n)\,a_n + c_1(n)\,a_{n-1} + \cdots + c_k(n)\,a_{n-k} = 0, $$ where each $c_i(n)$ is a polynomial (or constant) in $n$.

And generation functional meets next functional equation $$G[ g, \xi]= 8- G[ g/\xi , 1/\xi]$$ and differential equation of third order: \begin{eqnarray} g^3 \frac{\partial^3 G}{\partial g^3} +3 g^2 (\xi -1) \frac{\partial^3 G}{ \partial g^2 \partial \xi} +3 g (\xi-1) \xi \frac{\partial^3 G}{ \partial g\partial^2 \xi } +(\xi^2-1) \xi \frac{\partial^3 G}{\partial \xi^3}+ \nonumber \\ +6 \left( g^2 \frac{\partial^2 G}{\partial g^2} +g (2 \xi-1) \frac{\partial^2 G}{ \partial \xi \partial g} + \xi^2 \frac{\partial^2 G}{\partial \xi^2} + g \frac{ \partial G}{\partial g}+ \xi\frac{\partial G}{\partial \xi}\right) =0 \, . \end{eqnarray}

I have derived that $G(g, \xi)$ satisfies a third–order algebraic differential equation, but I was unable to extract from it a closed–form recurrence for $\{a_n\}$. Standard tools for guessing recurrences did not yield a stable result.

Answer 1

"Guessed", not proven, but validated for the first 80 coefficients:

$$(1 + 3n + 2n^2) a_n = -\frac{2((10 - 27n + 23n^2) + (-4 + 54n - 62n^2)\xi + (10 - 27n + 23n^2)\xi^2)}{(\xi+1)^3} a_{n-1} - \frac{48((26 - 27n + 7n^2) + (-88 + 90n - 22n^2)\xi + (26 - 27n + 7n^2)\xi^2)}{(\xi+1)^4} a_{n-2} - \frac{128(n-4)(5n-13)}{(\xi+1)^3} a_{n-3} + \frac{1024(n-5)(n-4)}{(\xi+1)^4} a_{n-4}$$

I first used resultants to eliminate $\bar g$, obtaining $$16g^2(\xi-1)^3G^3 + 8g(\xi-1)^2(-8g^2-20g\xi+\xi^2+28g+2\xi+1)G^2 + (\xi-1)(512g^3\xi+192g^2\xi^2+24g\xi^3+\xi^4-512g^3-1024g^2\xi-88g\xi^2+4\xi^3+704g^2+40g\xi+6\xi^2+152g+4\xi+1)G + 8(64g^2\xi+16g\xi^2+\xi^3-64g^2-112g\xi+3\xi^2+88g+3\xi+1) = 0$$

This allows expansion to obtain coefficients, and using an ansatz for the recurrence, the first $15$ coefficients suffice to extract the guess.

Answer 2

The question is about the power series expansion of $G(g, \xi) = \sum_{n \ge 0} a_n g^n$ where $G$ satisfies an algebraic differential equation. I examined the first few coefficients and found

$$ \frac18\frac{1\!+\!\xi}{1\!-\!\xi} G\left(\frac{1}{8}g(1\!+\!\xi)^3,\xi \right) \!=\! \frac{1}{(1\!-\!\xi)^2} \!-\! \frac{\xi^2}{(1\!-\!\xi)^2}g \!+\! \xi^2\sum_{k=2}^\infty b_k(\xi) g^k \tag1 $$

where $b_k(\xi)$ is a palindromic polynomial in $\xi$ with integer coefficients of degree $2(k-2)$.

$$ b_2 = 1,\: b_3 = -(1-4\xi+\xi^2),\: b_4 = 1-10\xi+22\xi^2-10\xi^3+\xi^4,\: \dots \tag2 $$

Based on the A.D.E., and using the code in my answer to https://mathematica.stackexchange.com/questions/285008), my calculations suggest that

$$ 0 = c_0\,b_k + c_1\,b_{k+1} + c_2\,b_{k+2} + c_3\,b_{k+3} \tag3 $$

for all $k\ge2$ with

\begin{align} c_0 &= -k(k-1)(3k+8)(1+\xi)^6, \\ c_1 &= +k(18k^2+75k+67)(1+\xi)^4, \\ c_2 &= +(3k+7)(80(1-4\xi+\xi^2)+(3k^2+13k)(5\xi^2-26\xi+5)), \\ c_3 &= +4(k+4)(2k+7)(3k+5). \tag4 \end{align}

Note carefully that the recurrence in equation $(3)$ is valid for $k=0$ or $k=1$ no matter what $b_0$ or $b_1$ is. This is because of the factor of $k$ in $c_1$ and the factor $k(k-1)$ in $c_0$.