Dtyjlui

So I am trying to find the sequence $(a_n)_{ngeq0}$ such that
$$frac1{x^2+x+1}=sum_{ngeq0}a_nx^n$$
My attempts:

I defined $$A(x)=frac1{x^2+x+1}$$
Hence $$A(x)=sum_{ngeq0}a_nx^n$$
And so $$(x^2+x+1)sum_{ngeq0}a_nx^n=1$$
$$sum_{ngeq0}a_nx^{n+2}+sum_{ngeq0}a_nx^{n+1}+sum_{ngeq0}a_nx^n=1$$
$$sum_{ngeq0}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$$a_0(x^2+x+1)+sum_{ngeq1}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$x=0$:
$$a_0=1$$
So
$$x^2+x+1+a_1(x^3+x^2+x)+sum_{ngeq2}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$$frac{x^2+x+1+a_1(x^3+x^2+x)+sum_{ngeq2}a_n(x^{n+2}+x^{n+1}+x^n)}x=frac1x$$
$$x+1+a_1(x^2+x+1)+sum_{ngeq2}a_n(x^{n+1}+x^{n}+x^{n-1})=0$$
$x=0$:
$$a_1=-1$$
One more time:
$$x^2+x+1-x^3-x^2-x+a_2(x^4+x^3+x^2)+sum_{ngeq3}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$$-x^3+a_2(x^4+x^3+x^2)+sum_{ngeq3}a_n(x^{n+2}+x^{n+1}+x^n)=0$$
Divide both sides by $x^2$:
$$-x+a_2(x^2+x+1)+sum_{ngeq3}a_n(x^{n}+x^{n-1}+x^{n-2})=0$$
$x=0$:
$$a_2=0$$
How do I find $a_n$? Am I doing things right so far? Thanks.

You could have continue your way.

Starting from
$$sum_{n=0}^infty a_nx^{n+2}+sum_{n=0}^infty a_nx^{n+1}+sum_{n=0}^infty a_nx^n=1$$ you already found $a_0=1$ and $a_1=-1$.

Now, consider a given degree $m >1$. You then have the recurrence relation
$$a_{m-2}+a_{m-1}+a_m=0$$ and the roots of the characteristic equation
$$r^2+r+1=0 implies r_pm=frac{-1 pm i sqrt 3} 2$$ making by the end (I let to you the intermediate steps)
$$a_m=cos left(frac{2 pi m}{3}right)-frac{1}{sqrt{3}}sin left(frac{2 pi
m}{3}right)$$ making $a_{3m}=1$, $a_{3m-1}=0$, $a_{3m+1}=-1$.

begin{align}
frac1{1+x+x^2} &= frac{1-x}{1-x^3} \
&= frac1{1-x^3}-xleft(frac1{1-x^3} right)\
&= sum_{i=0}^infty x^{3i}-xsum_{i=0}^infty x^{3i}\
&=sum_{i=0}^infty x^{3i}-sum_{i=0}^infty x^{3i+1}\
end{align}

Another way of doing this is to let the roots of $1+x+x^2=0$ be $a$ and $b$ so that $1+x+x^2=(1-ax)(1-bx)$ with $ab=1$ and $a+b=-1$. Then use partial fractions $$frac 1{1+x+x^2}=frac A{1-ax}+frac B{1-bx}$$ where clearing fractions gives $$1=A(1-bx)+B(1-ax)$$

Set $a=frac 1b$ to obtain $1=B(1-frac ab)$ so that $B=frac b{b-a}$ and similarly $A=frac a{a-b}$ so that $$frac 1{1+x+x^2}=sum_{n=0}^inftyfrac a{a-b}a^nx^n-sum_{n=0}^infty frac b{a-b}b^bx^n=sum_{n=0}^inftyfrac {a^{n+1}-b^{n+1}}{a-b}x^n$$

Now observe $a^2=-1-a=b$ and $a^3=ab=1$ and similarly $b^2=-1-b=a$ and $b^3=1$. For $n=3m$ the coefficient is $frac {a-b}{a-b}=1$, for $n=3m+1$ the coefficient is $frac {b-a}{a-b}=-1$ and for $n=3m+2$ the coefficient is $0$.

We can also identify $a=omega$ and $b=omega^2$ as being primitive cube roots of $1$. Note how the coefficients are periodic with period $3$. These facts are related, as another answer shows.

Note: I have tried to express this in such a way as to show how partial fractions can be used more generally - even where a convenient short cut is not available - it doesn't matter that you get complex roots, the answer ends up with real coefficients. The only slight trick is that we would normally factorise $$x^2+px+q=(x-c)(x-d)$$ but, setting $a=frac 1c, b= frac 1d$ we need the form $$x^2+px+q=cd(1- ax)(1-bx)=q(1-ax)(1-bx)$$In the case above we had $q=1$, but in general the factor must not be forgotten.