Proove that Abel polynomials are of binomial type There are many places where the statement that these polynomials are of binomial type but no one actually proves that
Let
$$
P_n(x) = x(x + an)^{n-1}
$$
Prove that
$$
P_n(x+y) = \sum_{k=0}^{n}C_n^{k}P_k(x)P_{n-k}(y),\\ \text{where}\ C_n^k=\frac{n!}{k!(n-k)!}
$$
 A: We seek to prove that
$$P_n(x+y) = \sum_{k=0}^n {n\choose k} P_k(x) P_{n-k}(y)$$
where
$$P_n(x) = x(x+an)^{n-1}$$
is an Abel polynomial. Introduce $T(z)$ with functional equation
$$T(z) = z \exp(a T(z))$$
Viewing this as an EGF we seek the coefficient
$$n! [z^n] \exp(x T(z)) =
x (n-1)! [z^{n-1}] \exp(x T(z)) T'(z).$$
Note that $[z^0] \exp(x T(z)) = 1.$
With the Cauchy Coefficient Formula we find for $n\ge 1$
$$\frac{x(n-1)!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n}
\exp(xT(z)) T'(z) \; dz.$$
Now we put $T(z)= w$ to get $z=w/\exp(aw)$ and
$$\frac{x(n-1)!}{2\pi i}
\int_{|w|=\gamma}
\frac{\exp(anw) \exp(xw)}{w^n} \; dw
\\ = \frac{x(n-1)!}{2\pi i}
\int_{|w|=\gamma}
\frac{\exp((x+an)w)}{w^n} \; dw
\\ = x (x+an)^{n-1}.$$
This means that
$$\exp(x T(z)) = 1 + \sum_{n\ge 1} x (x+an)^{n-1} \frac{z^n}{n!}
\\ = \sum_{n\ge 0} x (x+an)^{n-1} \frac{z^n}{n!}
= \sum_{n\ge 0} P_n(x) \frac{z^n}{n!}.$$
By convolution of EGFs we thus have
$$P_n(x+y) = n! [z^n] \exp((x+y) T(z))
= n! [z^n] \exp(xT(z)) \exp(yT(z))
\\ = n! \sum_{k=0}^n
[z^k] \exp(xT(z)) [z^{n-k}] \exp(yT(z))
\\ = n! \sum_{k=0}^n
\frac{P_k(x)}{k!} \frac{P_{n-k}(y)}{(n-k)!}
= \sum_{k=0}^n {n\choose k} P_k(x) P_{n-k}(y).$$
The CCF can also be done by Lagrange Inversion, which goes as follows.
Using the notation from Wikipedia on
Lagrange-Bürmann
we have $\phi(w) = \exp(aw)$ and $H(w) = \exp(xw)$ and we find
$$n! [z^n] \exp(x T(z)) = n! \frac{1}{n} [w^{n-1}]
x \exp(xw) \exp(anw)
\\ = (n-1)! x [w^{n-1}] \exp((x+an)w)
= x (x+an)^{n-1}.$$
