Suppose I have the recurrence formula, for $n\ge 1$,
$$\left(A+\frac{2}{3}\right)^n+w_3^2\left(A+\frac{1+w_3}{3}\right)^n+w_3\left(A+\frac{1+w_3^2}{3}\right)^n=0; A_0=1$$ where $w_3=\exp(2i\pi/3).$ Here the recurrence recurs once the terms in the parentheses are expanded and the exponents of $A$'s are dropped to subscripts. Hence, $$A^1=-\frac{1}{3}=A_1$$
Using the conversion fact that $1+w_3+w_3^2=0$, this becomes
$$\left(A+\frac{2}{3}\right)^n+w_3^2\left(A-\frac{w_3^2}{3}\right)^n+w_3\left(A-\frac{w_3}{3}\right)^n=0$$
Expanding the binomials gives
$$\sum_{k=0}^n\binom{n}{k}A_k\left(\frac{2}{3}\right)^{n-k}+\sum_{k=0}^n\binom{n}{k}w_3^2A_k(-1)^{n-k}\left(\frac{w_3^2}{3}\right)^{n-k}+\sum_{k=0}^n\binom{n}{k}w_3A_k(-1)^{n-k}\left(\frac{w_3}{3}\right)^{n-k}=0$$
$$\sum_{k=0}^n\binom{n}{k}\frac{2^{n-k}+(-1)^{n-k}w_3^{2(n-k+1)}+(-1)^{n-k}w_3^{n-k+1}}{3^{n-k}}A_k=0$$
Now I can peel off the $n$-th term which is also $0$ and get
$$\sum_{k=0}^{n-1}\binom{n}{k}\frac{2^{n-k}+(-1)^{n-k}w_3^{2(n-k+1)}+(-1)^{n-k}w_3^{n-k+1}}{3^{n-k}}A_k=0$$
Peeling off the $(n-1)$-th term then,
$$\sum_{k=0}^{n-2}\binom{n}{k}\frac{2^{n-k}+(-1)^{n-k}w_3^{2(n-k+1)}+(-1)^{n-k}w_3^{n-k+1}}{3^{n-k}}A_k+\binom{n}{n-1}\frac{2-w_3-w_3^2}{3}A_{n-1}=0$$
$$\sum_{k=0}^{n-2}\binom{n}{k}\frac{2^{n-k}+(-1)^{n-k}w_3^{2(n-k+1)}+(-1)^{n-k}w_3^{n-k+1}}{3^{n-k}}A_k+nA_{n-1}=0$$
This gives the then the recurrence relation
$$nA_{n-1}=-\sum_{k=0}^{n-2}\binom{n}{k}\frac{2^{n-k}+(-1)^{n-k}w_3^{2(n-k+1)}+(-1)^{n-k}w_3^{n-k+1}}{3^{n-k}}A_k$$
or, in terms of just $A_n$,
$$A_{n}=-\sum_{k=0}^{n-1}\binom{n+1}{k}\frac{2^{n+1-k}+(-1)^{n+1-k}w_3^{2(n-k+2)}+(-1)^{n+1-k}w_3^{n-k+2}}{(n+1)3^{n+1-k}}A_k$$
Now to get the E.G.F, i would multiply by $x^n/n!$ and running over the non-negative integers, setting that equal to $A(x)$, and so,
$$A(x)=\sum_{n=0}^\infty{A_n}\frac{x^n}{n!}$$
However, given the complexity of $A_n$, this seems like a disastrous approach. How can i ascertain the E.G.F in a much less computational manner. Is there a trick I'm missing that might make life a little easier finding $A(x)$?