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  • Let $f(n)$ be an arbitrary function such that $f(n)\in\mathbb{Z}$.
  • Let $a(n)$ be an integer sequence with generating function $\frac{1}{G(0)}$ where $$ G(j)=1-\frac{f(j)x}{G(j+1)} $$ Here we have $$ G(0)=1-\frac{f(0)x}{G(1)}=1-\frac{f(0)x}{1-\frac{f(1)x}{G(2)}}=1-\frac{f(0)x}{1-\frac{f(1)x}{1-\frac{f(2)x}{G(3)}}} $$ and so on.
  • Let $$ \begin{split} R_1(n,q) & =\sum\limits_{j=0}^{q+1}f(j)R_1(n-1,j),\\ R_1(0,q) & =[q=0] \end{split} $$ Here square brackets denote Iverson brackets.

I conjecture that $$R_1(n,0)=a(n).$$

Moreover, let $$ \begin{split} R_2(n,q) &= \sum\limits_{j=0}^{q+1}f(j)R_2(n-1,j), \\ R_2(0,q) &= z^q. \end{split} $$ Is there a way to determine $[z^k]R_2(n,0)$ using continued fractions?

More generally, let $g(n)$ be an arbitrary function such that $g(n)\in\mathbb{N}$, and let $$ \begin{split} R_3(n,q) &= \sum\limits_{j=0}^{g(q)}f(j)R_3(n-1,j), \\ R_3(0,q) &= z^q \end{split} $$ Is there a way to determine $[z^k]R_3(n,0)$ using continued fractions?

UPD: I think that the following observation can also be placed in this question:

  • Let $b(n)$ be an integer sequence with generating function $A(x)$ where $$ A(x)=1+\sum\limits_{n>0}f(n-1)(xA(x))^n $$
  • Let $$ \begin{split} R_4(n,q) & =R_4(n-1,q+1) + \sum\limits_{j=0}^{q}f(q-j)R_4(n-1,j),\\ R_4(0,q) & =[q=0] \end{split} $$

I conjecture that $$R_4(n,0)=b(n).$$

See for example A192206 and A204218. Memorizing the values of $R_4(n,q)$ speeds up the calculation of these sequences.

Is there a way to prove it? Are there positive answers to questions related to continued fractions?

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    $\begingroup$ The update seems to have a tidier generalisation. Let $F$ be the ogf of $f$ and $A$ satisfy $A(x) = F(xA(x))$. Conjecture: $[x^n] A(x) = R_5(n,0)$ where $$R_5(n,q) = \begin{cases}[q=0]f_0 & \textrm{if } n = 0 \\ \sum_{j=0}^{q+1} f_0^{q-j} f_{q-j+1} R_5(n-1,j) & \textrm{otherwise} \end{cases}$$ $\endgroup$
    – Peter Taylor
    Commented Jun 27, 2023 at 10:41
  • $\begingroup$ @PeterTaylor, thank you for comment! Nice conjecture! Could you share how you came to this conclusion? Are there non-recursive methods to compute $A(x)$? $\endgroup$
    – Notamathematician
    Commented Jun 27, 2023 at 18:07
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    $\begingroup$ I started by rephrasing your update in terms of $A(x) = F(xA(x))$ with $f_0 = 1$, and then wondered whether the restriction could be removed. Trial and error with Sage, tweaking the recurrence for $R_4$, gave a version of $R_5$ which fits for low terms. I'm trying to prove a non-recursive expression for the coefficients of $A$ as sums over lattice points in a hyperplane: $$A(x) \stackrel?= \sum_{n \ge 0} \frac{x^n}{n+1} \sum_{\substack{v \in \mathbb{N}^{n+1},\\ \sum_i v_i = n}} \prod_i f_{j_i}$$ $\endgroup$
    – Peter Taylor
    Commented Jun 28, 2023 at 7:43

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