TITLE: What is the coefficient of $x^{2k}$ in the $n$-th iterate, $f^{(n)}(x)$, if $f(x)=1+x^2$?
QUESTION [12 upvotes]: Let $f(x)=1+x^2$, and its $n$-th compositional iterate $f^{(n)}=\sum_{k=0}^N  c_{k,n} x^{2k}$, where $N=2^{n-1}$.  What do we know about $c_{k,n}$?

REPLY [3 votes]: There is some computation for iteration of $M(x) = x^2+c$ in the last section of G. Edgar, "Fractional iteration of series and transseries", to apear in Trans. Amer. Math. Soc. LINK 
Put $c=1$ in (19) to get
$$
M^{[s]} = x^{2^s}\left(
1+2^{-1+s}x^{-2}-2^{-1}x^{-2^{s+1}}+2^{-3+2s}x^{-4}
+2^{-2+s}x^{-2-2^{s+1}}
+O(x^{-6}+x^{-2^{s+2}})\right)
$$
for the $s$th iterate.  More terms could be done by the same method, just more computation.  The motivation there is fractional $s$, but it holds in particular for integer $s$.
So we have these:
$$\begin{align*}
M^{[0]} &= x+O(x^{-3})
\\
M^{[1]} &= x^2+1+O(x^{-4})
\\
M^{[2]} &= x^4+2x^2+2+O(x^{-2})
\\
M^{[3]} &= x^8+4x^6+8x^4+O(x^2)
\end{align*}$$
When $s$ is a positive integer, of course there are no negative powers.