TITLE: formula for the $k$th coefficient of a polynomial plugged into itself $n$ times
QUESTION [5 upvotes]: 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 $b_k$?

REPLY [3 votes]: Hint:
The first and second autocompositions of the quadratic polynomial are
$$a_2^3x^4+2a_1a_2^2x^3+(a_2a_1^2+2a_0a_2^2+a_1a_2)x^2+(2a_0a_1a_2+a_1^2)x+a_2a_0^2+a_0a_1+a_0$$
and
$$a_2^7x^8+4a_1a_2^6x^7+(6a_1^2a_2^5+4a_0x^6a_2^6+2a_1a_2^5x^6)+(12a_0a_1a_2^5+4a_1^3a_2^4+6a_1^2a_2^4)x^5+(a_2^3a_1^4+12a_0a_1^2a_2^4+a_1^2a_2^3+6a_0^2a_2^5+6a_0a_1a_2^4+6a_1^3a_2^3+2a_0a_2^4+a_1a_2^3)x^4+(4a_0a_1^3a_2^3+12a_1a_0^2a_2^4+12a_0a_1^2a_2^3+2a_2^2a_1^4+4a_0a_1a_2^3+2a_2^2a_1^3+2a_1^2a_2^2)x^3+ (6a_0^2a_1^2a_2^3+4a_0^3a_2^4+6a_0a_2^2a_1^3+6a_1a_0^2a_2^3+a_2a_1^4+4a_0a_1^2a_2^2+4a_0^2a_2^3+a_2a_1^3+4a_0a_1a_2^2+a_2a_1^2)x^2+(4a_1a_0^3a_2^3+6a_0^2a_1^2a_2^2+4a_1a_0^2a_2^2+2a_0a_2a_1^3+4a_0a_2a_1^2+a_1^3)x+a_2^3a_0^4+2a_1a_2^2a_0^3+a_2a_0^2a_1^2+2a_2^2a_0^3+3a_1a_2a_0^2+a_0a_1^2+a_2a_0^2+a_0a_1+a_0.$$
Appetizing, no ? I dare not think of the third composition of a quartic...