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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.

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  • $\begingroup$ I guess you can write it as the product of the generating functions oeis.org/A001590 and oeis.org/A005314, but I don't know whether this is elegant. $\endgroup$
    – Martin Rubey
    Commented Mar 31, 2022 at 15:03
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    $\begingroup$ I see from the answers that this is a stimulating question, but looking at the question alone I think that it badly fails the criterion of being well motivated. $\endgroup$
    – Dirk
    Commented Apr 1, 2022 at 4:34

3 Answers 3

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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.

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  • $\begingroup$ (this explains the comment I made above) $\endgroup$
    – Martin Rubey
    Commented Mar 31, 2022 at 15:50
  • $\begingroup$ Does one need to say "induction" is used to show monotonicity of $f(n)$ here? $\endgroup$
    – T. Amdeberhan
    Commented Mar 31, 2022 at 17:21
  • $\begingroup$ Yes, by "it follows" I meant "it follows by induction." $\endgroup$
    – Richard Stanley
    Commented Mar 31, 2022 at 18:43
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    $\begingroup$ On the lines of the proof for the other fraction, we may also write $$\frac{1-z}{1-2z+z^2-z^3}=\frac{1}{1- \big(z +\frac{z^3}{1-z}\big)}=\sum_{m\ge0}\Big(\sum_{k\ge1\atop k\neq2 } z^k\Big)^m $$ $\endgroup$
    – Pietro Majer
    Commented Mar 31, 2022 at 20:12
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    $\begingroup$ (oh this has already been remarked) $\endgroup$
    – Pietro Majer
    Commented Mar 31, 2022 at 20:18
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The rational function $(1-z)/(1-2z+z^2-z^3)$ has positive coefficients because it's equal to $$\frac{1}{1-z-\displaystyle\frac{z^3}{1-z}}.$$

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  • $\begingroup$ Thank you here also. $\endgroup$
    – T. Amdeberhan
    Commented Mar 31, 2022 at 20:00
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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.)

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  • $\begingroup$ I appreciate it. $\endgroup$
    – T. Amdeberhan
    Commented Mar 31, 2022 at 20:00

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