Every sixth polynomial shares a factor of $(a^2-6)$ I currently looking at the polynomials you get from the series expansion of 
$$
\frac{1-x^2}{1-ax+2x^2}=1+a x
 +(a^2-3) x^2
 +(a^3-5a) x^3
 +\underbrace{(a^4-7a^2+6)}_{(a-1)(a+1)(a^2-6)} x^4+\dots
$$
W|A helped here...
What I found is that from $x^{4}$ onwards, every sixth polynomial shares a factor of $(a^2-6)$. How to prove that?

Consider $\displaystyle f(a,x)=\frac{1-x^2}{1-a x+2x^2}+{1\over2}=\frac{3-ax}{2(1-a x+2x^2)}$, which is just your function augmented by $1/2$. What you ask amounts at proving that the series expansions of both $f(\sqrt6, x)$ and $f(-\sqrt6, x)$ around $x=0$ do not contain powers of $x$ with exponent ${4+6k}$. 
That is true because, by multiplying both numerator and denominator of $f(\pm\sqrt6, x)$ by $(1 \pm \sqrt6 x + 2 x^2)(1 + 2 x^2)$, one gets:
$$
f(\pm\sqrt6, x)
={3/2 \pm \sqrt6 x + 3 x^2 \pm \sqrt6 x^3 \mp 2 \sqrt6 x^5\over 1+8x^6}\\
=(3/2 \pm \sqrt6 x + 3 x^2 \pm \sqrt6 x^3 \mp 2 \sqrt6 x^5)
\sum_{k=0}^\infty(-8x^6)^k
$$
and the polynomial factor does not contain $x^4$.
