I have the following recurrence relation:
$G_0=0, G_1=1,$ $$\left(G+\frac{2}{3}\right)^n+\left(G+\frac{1+w_3}{3}\right)^n+\left(G+\frac{1+w_3^2}{3}\right)^n=0, n>1$$ where $w_3$ is the primitive $3$rd root of unity, and in the math of the Umbral Calculus, the exponents of G are brought down as subscripts once the expansion occurs.
For those who don't know how this works, take the Bernoulli numbers. They can be defined by the following recurrence:
$B_0=1, B_1=\frac{-1}{2}$ $$(B+1)^n-B^n=0, n>1$$
So take $n=3$. Then
$$(B+1)^3-B^3=0\Rightarrow B^3+3B^2+3B+1-B^3=0\Rightarrow 3B_2+3B_1+1=0$$
Thus since $B_1=-1/2$, we can solve for $B_2$.
$$3B_2-\frac{3}{2}+1=0\Rightarrow 3B_2=\frac{1}{2}\Rightarrow B_2=\frac{1}{6}$$
I'm not even sure I know where to begin. I can simply find a few of the coefficients using Mathematica or Maple, but how do I find the exponential generating function?