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?
 A: I will prove what you want kind of indirectly using Chebyshev polynomials of second kind. Actually once you formulate the question in terms of these polynomials this mysterious $a^2-6$ becomes quite meaningful. Let me start by  The generating function of Chebyshev polynomial of second kind, $U_n$ (which you can take as the definition of these polynomials):
$$
\frac{1}{1-2ax+x^2}=\sum_{n=0}^\infty U_n(a)x^n
$$
So your function which I will call $I(x,a)$ is nothing but
$$
I(x,a):=\frac{1-x^2}{1-ax +2x^2}\xrightarrow{t=\sqrt{2}x}\left[1-\frac{t^2}{2}\right]\sum_{n=0}^\infty U_n\left(\frac{a}{\sqrt{8}}\right)t^n$$
We are only interested in the value of this function when $a=\pm \sqrt{6}$. But then $\frac{a}{2\sqrt{2}}=\cos \pi/6$ or $=\cos(\pi-\pi/6)$ (You can already see where every sixth term pattern is coming from). Then your function becomes (I defined $X:=a/\sqrt{8}$)
$$
I(X,t):=U_0(X) + tU_1(X)+
t^2\sum_{n=0}^\infty \left[U_{n+2}(X)-\frac{1}{2}U_n(X)\right]t^n
$$
We are only interested in $n=6k+2$ and want to prove that
$$
2U_{4+6k}\left(\cos \frac{\pi}{6}\right)=
U_{2+6k}\left(\cos \frac{\pi}{6}\right)
$$
Now using yet another property of Chebyshev polynomial of second kind, in general if $z=\cos \psi$, then
$$
U_m(z) = \frac{\sin((m+1)\psi)}{\sin \psi}
$$
In our case then one can easily show with identity above that $
U_{4+6k}\left(\cos \frac{\pi}{6}\right)=
(-1)^k
$ and $
U_{2+6k}\left(\cos \frac{\pi}{6}\right)=2(-1)^k
$. This show that $a=+\sqrt{6}$ is a root for every sixth polynomial after $x^4$ in the expansion of $I(x,a)$. Similarly you can check the $a=-\sqrt{6}$ is also a root.
These identities about Chebyshev polynomials are surprisingly easy to prove actually. For a fast reference about Chebyshev polynomials, look here, or wikipedia.
A: 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 $x^4$.
A: Here is how you can answer a question like this in general, though I have not worked out the computation in this particular case.  A polynomial is divisible by $a^2-6$ iff it vanishes when you plug in $a=\pm\sqrt{6}$.  So your question really is: if $a=\pm\sqrt{6}$, does the coefficient of $x^{6n+4}$ in $f(x)=\frac{1-x^2}{1-ax+2x^2}$ vanish for each $n$?
To answer this, let's answer a much more general question.  Suppose you have a power series $f(x)=\sum c_n x^n$ (convergent in some neighborhood of $x=0$) and you want to know whether all the coeffients $c_n$ vanish when $n\equiv k$ mod $m$, for some fixed $k$ and $m$.  Let us write $f_j(x)=\sum c_{mn+j}x^{mn+j}$ for $j=0,1,\dots,m-1$; we want to know whether $f_k(x)$ is identically $0$.  To answer this, let $\zeta$ be a primitive $m$th root of $1$, and consider the functions $g_j(x)=f(\zeta^jx)$, for $j=0,1,\dots,m-1$.  Note that $g_j(x)=\sum_{\ell=0}^{m-1} \zeta^{j\ell} f_\ell(x)$.  I claim that $mf_k(x)=\sum_{j=0}^{m-1}\zeta^{-kj}g_j(x)$.  Indeed, we have $$\sum_{j=0}^{m-1}\zeta^{-kj}g_j(x)=\sum_j\sum_\ell \zeta^{-kj}\zeta^{j\ell}f_\ell(x)=\sum_\ell \left(\sum_{j=0}^{m-1} (\zeta^{\ell-k})^j\right) f_\ell(x).$$
Now $\zeta^{\ell-k}$ is an $m$th root of $1$, so the sum $\sum_{j=0}^{m-1} (\zeta^{\ell-k})^j$ vanishes unless $\zeta^{\ell-k}=1$, i.e. unless $\ell=k$, in which case the sum is $m$.  So we get $\sum_{j=0}^{m-1}\zeta^{-kj}g_j(x)=mf_k(x)$.
Thus we can conclude that $f_k(x)=0$ iff $\sum_{j=0}^{m-1}\zeta^{-kj}g_j(x)=\sum_{j=0}^{m-1}\zeta^{-kj}f(\zeta^jx)=0$.  This sum is one we can (in principle) compute directly from a formula for $f$.  For instance, in your case, with $m=6$, $k=4$, and $f(x)=\frac{1-x^2}{1-ax+2x^2}$, you get the following sum:
$$\frac{1-x^2}{1-ax+2x^2}+\zeta^2\frac{1-\zeta^2x^2}{1-a\zeta x+2\zeta^2x^2}+\zeta^4\frac{1-\zeta^4x^2}{1-a\zeta^2x+2\zeta^4x^2}+\frac{1-x^2}{1+ax+2x^2}+\zeta^2\frac{1-\zeta^2x^2}{1-a\zeta^4x+2\zeta^2x^2}+\zeta^4\frac{1-\zeta^4x^2}{1-a\zeta^5x+2\zeta^4x^2}.$$
Here $\zeta$ is a primitive $6$th root of $1$.  While this is long and nasty, in principle it shouldn't be too hard to expand it all out and see if it vanishes identically if $a=\pm\sqrt{6}$.
