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.



Positivity of a one-variable rational function
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.


