We are given the following generating function : $$G(x)=\frac{x}{1+x+x^2}$$
The question is to provide a closed formula for the sequence it determines.
I have no idea where to start. The denominator cannot be factored out as a product of two monomials with real coefficients. Any sort of help to solve this problem is welcome!
From $$G(x)=\frac{x}{1+x+x^2}=\frac{x(1-x)}{(1+x+x^2)(1-x)}=\frac{x-x^2}{1-x^3}$$ we see that the sequence is periodic mod $3$. By checking the first few terms, $$a_n=\begin{cases}0&\text{if }n\equiv 0\pmod 3,\\ 1&\text{if }n\equiv 1\pmod 3,\\ -1&\text{if }n\equiv 2\pmod 3.\\ \end{cases}$$
Another way:
Write $$\frac{x}{1+x+x^2}=\sum_{n=0}^\infty a_nx^n,$$
whence $$x=(1+x+x^2)\sum_{n=0}^\infty a_nx^n.$$
From here $a_0=0, a_1=1, a_n+a_{n-1}+a_{n-2}=0$.
If you want the Binet formula for this you can solve the matrix
Set up the partial fraction $$\frac{x}{1+x+x^2}=\frac{A}{1-ax}+\frac{B}{1-bx}$$ where $$a=\frac{-1+\sqrt{3}i}{2}$$ and $$ b=\frac{-1-\sqrt{3}i}{2}$$ so that $$A(1-bx)+B(1-ax)$$ The matrix becomes \begin{bmatrix} -b&-a&1\\[0.3em] 1&1&0 \end{bmatrix} or
\begin{bmatrix} \frac{1+\sqrt{3}i}{2}&\frac{1-\sqrt{3}i}{2}&1\\[0.3em] 1&1&0 \end{bmatrix} which in reduced echelon form is \begin{bmatrix} 1&0&\frac{-\sqrt{3}i}{3} \\[0.3em] 0&1&\frac{\sqrt{3}i}{3} \end{bmatrix}
The closed form then becomes $$a_n=\frac{-\sqrt{3}i}{3}\left(\frac{-1+\sqrt{3}i}{2}\right)^n+\frac{\sqrt{3}i}{3}\left(\frac{-1-\sqrt{3}i}{2}\right)^n$$
$a_0=0$ $a_1=1$, $a_2=-1$, $a_3=0$, $a_4=1$, $a_5=-1$ etc.
