# How to solve $F(z,q) = q + z F(z,q) + z^2 \big( F(z,q^2) \big)^2$?

Is there a method allowing us to solve the following functional equation ?

$$F(z,q) = q + z F(z,q) + z^2 \big( F(z,q^2) \big)^2$$

• What do you mean by solve? Commented Jul 4, 2017 at 12:47
• I would like to find F(z,q) Commented Jul 4, 2017 at 15:45
• Assuming that $F(z,q)$ is a power series in $z$ with coefficients polynomials in $q$, you can use the equation to determine the polynomials in $q$ one after another. Have you tried that? Commented Jul 4, 2017 at 19:21
• @Somos No, I haven't tried this yet. Actually $F(z,q)$ corresponds to a formal power series with terms $a_{n,k}z^nq^k$. A coefficient $a_{a,k}$ enumerates rooted planar binary trees of size $n$ with $k$ being a product of outdegrees. So, definitely I can assume this, but I don't know how to how to proceed further, in order to obtain a closed form of $F$. Commented Jul 4, 2017 at 21:20
• You can recursively define $$F(z,q;0)=q$$ and for $n\ge1$ $$F(z,q;n)=q+zF(z,q;n-1)+z^2(F(z,q^2;n-1))^2.$$ Easy to do with e.g. Mathematica. This converges to $F(z,q)$ as $n\to\infty$ in the sense the $z^m$-term no longer changes when $n>m$. It begins $$F(z,q)=q+qz+(q+q^4)z^2+(q+3q^4)z^3+(q+6q^4+2q^{10})z^4+(q+10q^4+10q^{10})z^5+\cdots$$ Undoubtedly you knew about this. Adding it just in case. Commented Jul 6, 2017 at 21:11

Is there a method allowing us to solve the following functional equation ? $$F(z,q) = q + z F(z,q) + z^2 \big( F(z,q^2) \big)^2$$
To solve the functional equation, assume we have a series expansion $$F(z,q) = \sum_{n=0}^\infty a_n(q)z^n.$$ The functional equation then forces $$a_0(q) = q$$, and for $$n>0$$ $$a_n(q) = a_{n-1}(q) + \sum_{k=0}^{n-2} a_k(q^2)a_{n-2-k}(q^2).$$ We can express this sequence of polynomials in $$q$$ as a finite sum $$a_n(q) = \sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n}{2k}b_k(q)$$ where $$\, b_0(q) = q,\; b_1(q) = q^4,\; b_2(q) = 2q^{10},\; \dots\;$$ are another sequence of polynomials in $$q$$ with positive integer coefficients whose polynomial expansion looks like $$b_k(q) = u_k q^{A073121(k+1)} + \cdots + v_k q^{3\cdot 2^k-2}$$ where A073121 is a sequence in OEIS. I think it is unlikely that $$F$$ has a closed form.