<s>[INST] To prove that $a_n=[x^n]\frac {x-2x^2+x^3}{1-2x+x^2-x^4}$ To prove: $$a_n=[x^n]\frac {x-2x^2+x^3}{1-2x+x^2-x^4}$$
Given that, $a_n$is the number of compositions of $n$ in which the number of parts is odd with the first part being equal to the number of parts.
 [/INST] For $k\geq 1$ and $n\geq 3$, the number of positive integer solutions of
$$(2k+1)+x_2+x_3+\dots+x_{2k+1}=n$$
that is of
$$x_2+x_3+\dots+x_{2k+1}=n-(2k+1)$$
which is equal to
$$\binom{n-(2k+1)-1}{2k-1}.$$
Hence
$$a_n=\sum_{1\leq k\leq \lfloor (n-1)/4\rfloor}\binom{n-(2k-1)-1}{2k-1}.$$
Now by partial fraction decomposition
$$\frac{x-2x^2+x^3}{1-2x+x^2-x^4}=
\frac{1}{2(1-x+x^2)}-\frac{1-2x}{2(1-x-x^2)}$$
and 
$$\begin{align}[x^n]\frac{1}{1-x+x^2}&=[x^n]\sum_{m=0}^{\infty}(x-x^2)^m=[x^n]\sum_{m=0}^{\infty}
\sum_{j=0}^m(-1)^j\binom{m}{j}x^{m+j}=\sum_{0\leq j\leq \lfloor n/2\rfloor}
(-1)^j\binom{n-j}{j},\\
[x^n]\frac{1}{1-x-x^2}&=[x^n]\sum_{m=0}^{\infty}(x+x^2)^m=[x^n]\sum_{m=0}^{\infty}
\sum_{j=0}^m\binom{m}{j}x^{m+j}=\sum_{0\leq j\leq \lfloor n/2\rfloor}\binom{n-j}{j}.
\end{align}$$
Can you take it from here?
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