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How to probe the recursiveness order of a sequence $\{S_n\}$ whose generating function is known:

$$ \sum_{n\geq0} S_n z^n= \frac{4 z \left(\sqrt{49 z^2-18 z+1}+7 z-1\right)}{\sqrt{49 z^2-18 z+1} \left(\sqrt{49 z^2-18 z+1}+15 z-1\right) \left(3 \sqrt{49 z^2-18 z+1}+21 z-1\right)}+\frac{3}{3-44 z} $$

And even more, as the following is proven.

$ \textbf{Conjecture}$ : $\{S_n\}$ satisfy the following recurrence

$$ 0= P_0(n) S_n+ P_1(n) S_{n-1}+ P_2(n) S_{n-2}+ P_3(n) S_{n-3}+ P_4(n) S_{n-4}$$

where $P_i(n) \in \mathbb{R}[n]$ are polynomials of degree 2.

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    $\begingroup$ It is a differential equation for the function (of order 4, with corresponding polynomial coefficients). You may check whether it is satisfied by direct differentiating. $\endgroup$
    – Fedor Petrov
    Commented Jun 15, 2021 at 20:24
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    $\begingroup$ ... and in this case, unless I'm mistaken, the recurrence is $$\left(12 n^{2}-1581 n +1569\right) S_{n}+\left(-572 n^{2}+75469 n -111467\right) S_{n -1}+\left(9636 n^{2}-1273335 n +2500019\right) S_{n -2}+\left(-64964 n^{2}+8600211 n -21072477\right) S_{n -3}+\left(129360 n^{2}-17172540 n +50353380\right) S_{n -4} = 0 $$ $\endgroup$
    – Robert Israel
    Commented Jun 16, 2021 at 5:02
  • $\begingroup$ math.stackexchange.com/questions/4173685/… $\endgroup$
    – José María Grau Ribas
    Commented Jun 16, 2021 at 6:27
  • $\begingroup$ @Fedor Petrov Why do you know a priori that the order of the differential equation is 4. $\endgroup$
    – José María Grau Ribas
    Commented Jun 17, 2021 at 18:28
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    $\begingroup$ Sorry, of order 2, I misread the question. That's because the degrees of polynomials do not exceed 2 $\endgroup$
    – Fedor Petrov
    Commented Jun 17, 2021 at 18:49

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Extended comment.

The relation $$0=P_0(n) S_n+ P_1(n) S_{n-1}+ P_2(n) S_{n-2}+ P_3(n) S_{n-3}+ P_4(n) S_{n-4}$$ for all $n\geqslant 4$ is equivalent to $$ 0=\sum_{n\geqslant 4} (P_0(n) S_n+ P_1(n) S_{n-1}+ P_2(n) S_{n-2}+ P_3(n) S_{n-3}+ P_4(n) S_{n-4})x^n. $$ Denote $\sum_{n=0}^\infty S_n x^n=f(x)$. Then $f'(x)=\sum_{n\geqslant 0} S_n\cdot nx^{n-1}$, $f''(x)=\sum_{n\geqslant 0} S_n\cdot n(n-1)x^{n-2}$. Thus for $i\in \{0,1,2,3,4\}$ we have $$\sum_{n\geqslant 4} P_i(n)S_{n-i}x^n= \sum_{n\geqslant i} P_i(n)S_{n-i}x^n-\sum_{n=i}^3P_i(n)S_{n-i}x^i=\\ \sum_{n\geqslant 0} P_i(n+i)S_{n}x^{n+i}-\sum_{n=i}^3P_i(n)S_{n-i}x^i.$$ Denote $P_i(n+i)=a_i+b_in+c_in(n-1)$ and $\sum_{n=i}^3P_i(n)S_{n-i}x^i=g_i(x)$. Then $$\sum_{n\geqslant 0} P_i(n+i)S_{n}x^{n+i}-\sum_{n=i}^3P_i(n)S_{n-i}x^i=x^i(a_if(x)+b_ixf'(x)+c_ix^2f''(x))-g_i(x).$$ Thus the differential equation for $f$ has a form $$ q_0(x)f(x)+xq_1(x)f'(x)+x^2q_2(x)f''(x)=A(x) $$ for some polynomials $q_0,q_1,q_2$ of degree at most 4 and $A$ of degree at most 3.

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