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  • Let $f(n)$ be an arbitrary function such that $f(n)\in\mathbb{Z}$.
  • Let $$ F(x)=\sum\limits_{m\geqslant 0}f(m)x^m $$
  • Define the operator $\operatorname{SR}$, which is associated with the series reversion.
  • Let $a(n)$ be an integer sequence with generating function $A(x)$ where $$ A(x)=\frac{1}{x}\operatorname{SR}(xF(x)) $$
  • Let $b(n)$ be an integer sequence with generating function $\frac{1}{G(0)}$ where $$ G(j)=F\left(\frac{x}{G(j+1)}\right) $$ Here we have $$ G(0)=F\left(\frac{x}{G(1)}\right)=F\left(\frac{x}{F\left(\frac{x}{G(2)}\right)}\right)=F\left(\frac{x}{F\left(\frac{x}{F\left(\frac{x}{G(3)}\right)}\right)}\right) $$ and so on.

I conjecture that $$a(n)=b(n).$$

Is there a way to prove it?

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    $\begingroup$ Please give a simple example of $f(n)$ and $a_n$ in your question to motivate it. Perhaps from the OEIS. $\endgroup$
    – Somos
    Commented Jun 28, 2023 at 19:46
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    $\begingroup$ @Somos, I think the motivation is mathoverflow.net/q/449536/46140 and some discussion in the comments there (part of which has been deleted). Notamathematician, why not introduce $F$ as the generating function of $f$ and say $$A(x) = x^{-1} \operatorname{SR}(xF(x)) \\ G(j) = F(xG(j+1)^{-1})$$? $\endgroup$
    – Peter Taylor
    Commented Jun 29, 2023 at 7:14
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    $\begingroup$ @PeterTaylor I am still waiting for a simple example of $\,f(n)\,$ and $\,a_n.$ $\endgroup$
    – Somos
    Commented Jul 11, 2023 at 22:42

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We assume $F(0) \neq 0$, since otherwise we don't satisfy the assumptions for the series reversion. Let $G = G(0)$ be the fixpoint of the recurrence given:

$$G(x) = F\left(\frac{x}{G(x)}\right)$$

Multiply both sides by $\frac{x}{G(x)}$:

$$x = \frac{x}{G(x)} F\left(\frac{x}{G(x)}\right)$$

By inspection of the structure we have $\frac{x}{G(x)} = \operatorname{SR}(xF(x))$, or $$\frac{1}{G(x)} = \frac{1}{x} \operatorname{SR}(xF(x))$$ as desired.

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  • $\begingroup$ Thank you for answer! Could you explain how you got the result after "by inspection of the structure"? What is the general rule here? $\endgroup$
    – Notamathematician
    Commented Jun 30, 2023 at 8:05
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    $\begingroup$ The definition of series reversion is that $h(\operatorname{SR}(h)(x)) = x$. The LHS is $x$ and the RHS has the form $g(x)F(g(x))$. $\endgroup$
    – Peter Taylor
    Commented Jun 30, 2023 at 8:07
  • $\begingroup$ It means that we have $H(x)=xF(x), H(\operatorname{SR}(H(n)))=x=H(g(x)), \operatorname{SR}(H(n))=g(x)$, right? $\endgroup$
    – Notamathematician
    Commented Jun 30, 2023 at 9:33
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    $\begingroup$ $x=H(g(x)) \implies \operatorname{SR}(H(x))=g(x)$ $\endgroup$
    – Peter Taylor
    Commented Jun 30, 2023 at 10:00
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    $\begingroup$ It wasn't, but I only post the clean route to the solution rather than the dead ends I tried. $\endgroup$
    – Peter Taylor
    Commented Jun 30, 2023 at 10:46

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