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Define an operator $L$ on, say, formal series $f(x)$ with $f(0)=1$ by requiring that $L(f)=F$ is the solution of the functional equation $$ F(xf(x))=f(x). $$

Some examples: \begin{align*} L(1)&=1;\\ L\left(\frac1{1+cx}\right)&=1-cx;\\ L(1-cx)&=\frac{1+\sqrt{1-4cx}}2=1-\sum_{n=0}^\infty C_n(cx)^{n+1} \end{align*} where $C_n$ are the Catalan numbers; \begin{align*} L\left(\frac{1+\sqrt{1-4x}}2\right)&=\frac{1}{3} \left(1+2 \cos \left(\frac{1}{3} \arccos\left(1-\frac{27 x}{2}\right)\right)\right)\\&=1-x-2x^2-7x^3-...-\frac{\binom{3n+1}n}{n+1}x^{n+1}-...\\ &=\text{ solution of $F(x)^3-F(x)^2+x=0$} \end{align*} (see e. g. A006013 on OEIS);

\begin{align*} L\left(\frac1{(1+ax)(1+bx)}\right)&=\frac{1-(a+b)x+\sqrt{1-2(a+b)x+((a-b)x)^2}}2\\ &=1-(a+b)x-\sum_{n>0}\sum_{k=1}^n\frac1n\binom nk\binom n{k-1}a^kb^{n+1-k}x^{n+1} \end{align*} (Narayana numbers);

$$ L\left(\frac1x\ln\frac1{1-x}\right)=\frac{xe^x}{e^x-1}=1+\frac x2+\frac{x^2}{12}-\frac{x^4}{720}+...+\frac{B_nx^n}{n!}+... $$ with $B_n$ the Bernoulli numbers; $$ L(e^x)=e^{W(x)} $$ where $W(x)$ is the Lambert function.

Have the properties of this (or similar) operator been studied before?

Specifically I would like to find out what can be said about the value of this operator on the series $$ \sum_{n\geqslant0}(C_nx^n)^2=\frac1{4x^2}\left({}_2F_1\left(-\frac12,-\frac12;1,16x^2\right)-1\right) $$ with $C_n$ the Catalan numbers. This is the generating function of irreducible meander systems as computed in

Lando, S. K.; Zvonkin, A. K., Meanders, Sel. Math. Sov. 11, No. 2, 117-144 (1992). ZBL0792.05006.

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    $\begingroup$ oeis.org/A168344 $\endgroup$ Commented Jul 25 at 7:56
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    $\begingroup$ experimentally, if $f = 1 + \sum_{k>0} c_k x^k$, then apparently $[x^n] F(x)$ is a polynomial in $c_1,\dots, c_n$ with monomials of the form $c_\lambda$ for a partition $\lambda\vdash n$, whose sign is $(-1)^{\ell(\lambda)}$. Do the coefficients have a combinatorial interpretation? The coefficient of $c_1^n$ is the Catalan number. $\endgroup$ Commented Jul 25 at 9:59
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    $\begingroup$ The coefficients of $c_2^k c_1^{n-2k}$ for $k=1$ and $k=2$ also have nice interpretations in the oeis. $\endgroup$ Commented Jul 25 at 10:02
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    $\begingroup$ @MartinRubey, I think partition $\lambda = 1^{e_1} 2^{e_2} \cdots$ has coefficient $-(-n)^{\underline{\ell(\lambda)-1}} (e_1! e_2! \cdots)^{-1}$. I think it should be possible to prove this by induction. $\endgroup$ Commented Jul 25 at 11:27
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    $\begingroup$ Note that Bürmann-Lagrange gives in the situation above $$ [z^0] F(z)=1,\,[z^1] F(z)= [t^1] \big(\log f(t)\big) = [t^1] f(t)\mbox{ and }$$ $$ [z^n] F(z) = -\frac{1}{n-1} [t^{n-1}]\big(f(t)\big)^{-(n-1)} \mbox{ for }n\geq 2$$ $\endgroup$ Commented Jul 25 at 20:50

2 Answers 2

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These are the partition polynomials of OEIS A350499, the free moment partition polynomials determining the free cumulants from the free moments of free probability theory. They are refinements (mod signs) of OEIS A060693 and A088617, through which you can find additional associations to many other important OEIS arrays and some historical details.

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  • $\begingroup$ See mathoverflow.net/questions/412573/… for these partition polynomials and related analysis. $\endgroup$ Commented Jul 27 at 17:41
  • $\begingroup$ See also mathoverflow.net/questions/462034/… for the relation to compositional inversion of Laurent series. The comments in my numerous linked MO-Q&A note the appearance of these polynomials in several research papers. $\endgroup$ Commented Jul 27 at 17:55
  • $\begingroup$ See also mathoverflow.net/questions/422539/… for a broader algebraic context in which these partition polynomials appear. $\endgroup$ Commented Jul 30 at 16:57
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Thanks to the comments by Peter Taylor, Martin Rubey and esg I now have a reasonably satisfactory answer.

Let $x=g(z)$ be the solution of $z=xf(x)$, that is, the inversion of the series $xf(x)$. Then $$ F(z)=z/g(z) $$ (simply by $F(z)=F(xf(x))=f(x)=(xf(x))/x=z/g(z)$). Then by Bürmann-Lagrange we can explicitly compute coefficients for $F(z)$. If $f(x)=1-c_1x-c_2x^2-...$ and $F(z)=1-C_1z-C_2z^2-...$ then $$ C_n=\sum_{\substack{m_1+2m_2+...+nm_n=n}\\m_1,m_2,...,m_n\geqslant0}\frac{(n+m_1+...+m_n-2)!}{(n-1)!}\frac{c_1^{m_1}}{m_1!}\frac{c_2^{m_2}}{m_2!}\cdots\frac{c_n^{m_n}}{m_n!}. $$ Note that all the involved numbers are in fact integers. For example, \begin{align*} C_1&=c_1\\ C_2&=c_2+c_1^2\\ C_3&=c_3+3c_1c_2+2c_1^3\\ C_4&=c_4+4c_1c_3+2c_2^2+10c_1^2c_2+5c_1^4\\ C_5&=c_5+5(c_1c_4+c_2c_3)+15(c_1^2c_3+c_1c_2^2)+35c_1^3c_2+14c_1^5\\ C_6&=c_6+6(c_1c_5+c_2c_4)+3c_3^2+21c_1^2c_4+42c_1c_2c_3+7c_2^3+56c_1^3c_3+84c_1^2c_2^2+126c_1^4c_2+42c_1^6 \end{align*}

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