Question: 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?

Answer: 
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.