<s>[INST] Expansion of $x^4\over1+x^2$ into a power series I calculated the generating function $A(x)$ of the recurrence $a_n = a_{n-2} - 2a_{n-3}$, $(n \ge 0, a_0 = a_1 = 0, a_2 = 2)$ and I have no clue on how to expand it into a power series in order to read the coefficients.
$$A(x) = {-2x^4 \over 1+x^2}.$$
 [/INST] Your generating function is incorrect.
You have the recurrence $$a_n=a_{n-2}-2a_{n-3}+2[n=2]\;,$$ where the last term is an Iverson bracket, and I assume that $a_n=0$ for $n<0$. Then 
$$\begin{align*}
A(x)&=\sum_{n\ge 0}a_nx^n\\
&=\sum_{n\ge 0}a_{n-2}x^n-2\sum_{n\ge 0}a_{n-3}x^n+2x^2\\
&=x^2A(x)-2x^3A(x)+2x^2\;,
\end{align*}$$
so $$A(x)=\frac{2x^2}{1-x^2+2x^3}\;.$$
Your function is
$$\begin{align*}
\frac{-2x^4}{1+x^2}&=-2x^4\cdot\frac1{1-(-x^2)}\\
&=-2x^4\sum_{n\ge 0}(-x^2)^n\\
&=-2x^4\sum_{n\ge 0}(-1)^nx^{2n}\;,
\end{align*}$$
in which the coefficient of $x^2$ and of every odd power of $x$ is $0$. However, $a_2=2\ne 0$, and $a_3=a_1-2a_0=0$, so $a_5=a_3-2a_2=-4\ne 0$.
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