formula for the $k$th coefficient of a polynomial plugged into itself $n$ times If a polynomial with coefficients $a_k$ is plugged into itself $n$ times, this will result in another polynomial with polynomial coefficients $b_k$. Find an explicit formula for $b_k$ given $a_k$, and $n$.
For example, $f(x)$ is a polynomial with the following coefficients $a_0=1$, $a_1=2$, and $a_2=3$. All other coefficients $a_m$ where $m>2$ are zero. This will result in the following polynomial.
$$
f(t) = 1 + 2x + 3x^2
$$
We want to find the polynomial coefficients of this polynomial after it is plugged into itself 1 time, thus $n=1$.
$$
f(f(t)) = 1 + 2(1 + 2x + 3x^2) + 3(1 + 2x + 3x^2)^2
$$
$$
f(f(t)) = 6 + 16 x + 36 x^2 + 36 x^3 + 27 x^4
$$
Thus $b_0=6$, $b_1=16$, $b_2=36$, and so on.
So, in summary, if a polynomial with polynomial coefficients $a_k$ plugged into itself $n$ times, what are the coefficients of the resulting polynomial $b_k$?
 A: As Darius Chitu noted, the general problem looks complicated and a formula for $b_k$ can be very cumbersome.
Maybe a less complicated way is the following. The polynomial $f^{(n)}(t)$ has degree $d=(n+1)\deg f$, so it is determined by its values  at any $d+1$ points, for instance, at points $0,\dots, d$. We can easily calculate these values recursively, and then apply   Lagrange’s interpolation formula. It yields
$$f^{(n)}(t)=\sum_{i=0}^d f^{(n)}(i)\prod_{j=0,\, j\ne i}^d\frac {t-j}{i-j}=
\sum_{i=0}^d \frac {(-1)^{d-i}f^{(n)}(i)}{i!(d-i)!} \prod_{j=0,\, j\ne i}^d (t-j).$$
A: It's not particularly clear what sort of answer you're after. Taking the composition of generating functions is an extremely common operation in combinatorics, and one can frequently give such things combinatorial meaning. However, numerically it's just reinterpreting the naive description of $f(f(x)) = \sum_{k=0}^d a_k f(x)^d$ where $f(x)^d$ is expanded naively by picking one factor from each $f(x)$, multiplying $d$ of them, and summing.
A lovely exposition of this sort of thing is in Richard Stanley's Enumerative Combinatorics, Volume 2, section 5.1. It's cast in terms of exponential generating functions, but there's an ordinary generating function analogue. (It seems silly, but I don't have a handy to-the-point reference for the ordinary version, aside from my own course notes from when I last taught combinatorics. Sorry the file is enormous, see Lecture 25 on p.200.)
