**Description**

I my work, I met fixed-point functional equations of a very specific form. Since I am not expert in the domain of functional equations, I have not pushed the study of these very far. Here is a description the objects.

Let $X := \lbrace x_1, \dots, x_k \rbrace$ be an alphabet of mutually commuting indeterminates and $F(x_1, \dots, x_k)$ be the formal power series
\begin{equation}
    F(x_1, \dots, x_k) := 
    \sum_{\alpha_1, \dots, \alpha_k} 
    \lambda_{\alpha_1 \dots \alpha_k} \: 
    x_1^{\alpha_1} \dots x_k^{\alpha_k},
\end{equation}
where all coefficients $\lambda_{\alpha_1 \dots \alpha_k}$ are nonnegative integers.

The series $F(x_1, \dots, x_k)$ is defined by the fixed-point functional equation
\begin{equation}
    F(x_1, \dots, x_k) = x_1 + F(P_1(x_1, \dots, x_k), \dots, P_k(x_1, \dots, x_k)),
\end{equation}
where for any $i \in [k]$, $P_i(x_1, \dots, x_k)$ is a polynomial with nonnegative integer coefficients. Of course, there are some necessarily conditions to ensure that $F(x_1, \dots, x_k)$ is well-defined.

Here are two celebrated examples of fixed-point equations of this kind.

 1. The series $S(x)$ defined by
\begin{equation}
    S(x) = x + S(x^2 + x^3).
\end{equation}
This functional equation has been studied by A. M. Odlyzko (see [Odl82]). One can compute first coefficients of $S(x)$ by *iteration*:
\begin{equation}
    S_1(x) = x,
\end{equation}
\begin{equation}
    S_2(x) = x + x^2 + x^3,
\end{equation}
\begin{equation}
    S_3(x) = x + x^2 + x^3 + x^4 + 2x^5 + 3x^7 + 3x^8 + x^9,
\end{equation}
and so on. This generating series is of the form
\begin{equation}
    S(x) = x + x^2 + x^3 + x^4 + 2x^5 + 2x^6 + 3x^7 + 4x^8 + 5x^9 + 
    8x^{10} + 14x^{11} + 23x^{12} + \dots
\end{equation}
and its coefficients have a combinatorial interpretation: the coefficient of $x^n$ is the number of *$2,3$-balanced trees* (see [Odl82] and also [A014535][1]) with $n$ leaves. 

 2. The series $B(x, y)$ defined by
\begin{equation}
    B(x, y) = x + B(x^2 + 2xy, x).
\end{equation}
By iteration, one finds
\begin{equation}
    B_1(x, y) = x,
\end{equation}
\begin{equation}
    B_2(x, y) = x + 2xy + x^2,
\end{equation}
\begin{equation}
    B_3(x, y) = x + 2xy + x^2 + 4x^2y + 2x^3 + 4x^2y^2 + 4x^3y + x^4.
\end{equation}
The coefficient of $x^n$ is the specialization $B(x) := B(x, 0)$ is the number of *balanced binary trees* (see [BLL94] and also [A006265][2]) with $n$ leaves.

In my work, I encountered fixed-point functional equations rather more complicated:
\begin{equation}
    F(x, y, z) = x + F(x^2 + xy + yz, x, xy),
\end{equation}
\begin{equation}
    G(x, y, z) = x + G(x^2 + 2xy + yz, x, x^2 + xy),
\end{equation}
\begin{equation}
    H(x, y, z, t) = x + H(x^2 + 2yt + yz, x, x^2 + xy, yt + yz).
\end{equation}
Coefficients of these count some tree-like structures.

It seems that the theory is fairly well-known when the generating series are univariate (for instance, A. M. Odlyzko [Odl82] gives asymptotic formulas and recurrence relations for the coefficients of $S(x)$). In the multivariate case, it seems that only few things are known and up to my knowledge, there are no other way to express the series $B(x)$ counting balanced binary trees.

**Questions**

> (1) Have you encountered other fixed-point functional equations of the described form?

Given a fixed-point functional equation for a series $F$ of the given form:

> (2) What about an other method to **extract coefficients** of $F$ otherwise than iteration?

> (3) What about **recurrence relations** between the coefficients of $F$?

> (4) What about **asymptotic approximations** for the coefficients of $F$?

> (5) What about **rationality**, **algebraicness**, **d-finiteness**, or other of $F$?

**References**

[Odl82] A. M. Odlyzko. Periodic Oscillations of Coefficients of Power Series That Satisfy Functional Equations. *Advances in Mathematics*, 44:180–205, 1982

[BLL94] F. Bergeron, G. Labelle, and P. Leroux. *Combinatorial Species and Tree-like Structures*. Cambridge University Press, 1994.


  [1]: http://oeis.org/A014535
  [2]: http://oeis.org/A006265