Let $p$ be a power series with integer coefficients of the special form $p(z)=z+a_2z^2+a_3z^3+..$. I wonder if the inverse (composition not $1/p$) series has again integer coefficients.
I have calculated some of such series so I guess yes. What do you think?
On
Given $$ f(x)=x+a_2x^2+a_3x^3+\dots\tag{1} $$ and $$ g(x)=x+b_2x^2+b_3x^3+\dots\tag{2} $$ formally, for $n\ge2$, the coefficient of $x^n$ in $g\circ f$ is $$ 0=a_n+b_n+\left[x^n\right]\sum_{k=2}^{n-1}b_k\left(x+a_2x^2+a_3x^3+\dots+a_{n-1}x^{n-1}\right)^k\tag{3} $$ For $n=2$, $(3)$ says that $b_2=-a_2$. Then inductively, since $a_n\in\mathbb{Z}$, $(3)$ guarantees that $b_n\in\mathbb{Z}$.
Even if we only know for $2\le k\le n$, that $a_k\in\mathbb{Z}$, $(3)$ guarantees that for $2\le k\le n$, we have $b_k\in\mathbb{Z}$.
For example:
$b_2=-a_2$
$b_3=2a_2^2-a_3$
$b_4=5a_2a_3-5a_2^2-a_4$
$b_5=14a_2^4-21a_2^2a_3+3a_3^2+6a_2a_4-a_5$
Yes it is correct, either you assume
1) the original series converges in some where as a Taylor series; or 2) just "formal" calculations.
The key is you can list out a series of equations
b_1a_1=1
b_n + b_{n-1}a_1 + .. = 0 (n>1)
And you can decide one by one b_n.