To prove that $a_n=[x^n]\frac {x-2x^2+x^3}{1-2x+x^2-x^4}$

Question

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.

Answer 1

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?