Positivity of a one-variable rational function Let's consider the $1$-variable rational function
$$F(z):=\frac{1-z}{(z^3 - z^2 + 2z - 1)\,(z^3 + z^2 + z - 1)}.$$
Numerical evidence convinces me of the truth of the following.

QUESTION. Can you prove that $F(z)$ is positive, in the sense that its Taylor series at $z=0$ has positive coefficients?

Note. I'm not sure whether the concept of multi-sections of a series is efficient for the present purpose. Nor do I think that looking at asymptotic growth of largest positive real roots is any more elegant.
 A: Let $1/(1-2z+z^2-z^3)=\sum_{n\geq 0} f(n)z^n$. Then $f(0)=1$,
$f(1)=2$, $f(2)=3$, and
$f(n+1)=2f(n)-f(n-1)+f(n-2)=f(n)+(f(n)-f(n-1))+f(n-2)$, $n>2$. It
follows that $f(n)$ is strictly increasing, so $(1-z)/(1-2z+z^2-z^3)$
has positive coefficients. Similarly, or because
$$ \frac{1}{1-z-z^2-z^3} = \sum_{m\geq 0}(z+z^2+z^3)^m, $$
the series $1/(1-z-z^2-z^3)$ has positive coefficients. Hence the
product
$$ \frac{1-z}{1-2z+z^2-z^3}\cdot \frac{1}{1-z-z^2-z^3} $$
has positive coefficients.
A: The rational function $(1-z)/(1-2z+z^2-z^3)$ has positive coefficients because it's equal to
$$\frac{1}{1-z-\displaystyle\frac{z^3}{1-z}}.$$
A: One can show that all coefficients with sufficiently large index are positive. Indeed, using Maple, the pole of $f$ closest to the origin is:
$a:=0.543689...>0,$
and the residue at this pole is $c:=-0.3115580216...<0$. So
$$\frac{c}{z-a}=-\frac{c}{a}\sum_{n=0}^\infty \left(\frac{z}{a}\right)^n$$
has positive coefficients.
It is not difficult to estimate the integer $n_0$ such that for $n>n_0$
this part of the partial fraction decomposition will dominate the rest,
and the first $n_0$ coefficients can be computed using Maple.
(The second pole closest to the origin is $a_1=0.56984...$ and the residue
at it $c_1=0.3383$. So the contribution from $a$ overtakes
the contribution from $a_1$ already for $n\geq 2$.
The other 4 poles are two complex conjugate pairs,
and their absolute values are $>1$, and the residues less than 9 by absolute value, so they have no influence, say for $n>10$.
On the other hand Maple computes the first 100 or 200
coefficients in no time, and they are all positive.)
