A question on linear recurrence

Question

Let $(a_{i})$ be an increasing sequence of positive integers given by a linear recurrence $a_{i+n}=c_{n}a_{i+n-1}+\dots +c_{1}a_{i}$ with $c_{i}\in\{-1,0,1\}$ and $a_{i}=2^{i}$ for $i=1,\dots n$ such that the characteristic polynomial $p(x)=-x^{n}+c_{n}x^{n-1}+\dots +c_{1}x^{0}$ has a unique dominating real root $\alpha> 1$. Let $(s_{i})\in\{-1,0,1\}^{m}$ with $m\ge n$. Is it true that $\sum_{i=1}^{m}s_{i}a_{i}=0$ if and only if $p$ divides the polynomial $\sum_{i=1}^{m}s_{i}x^{i}$? Where I can find a proof of this kind of result?

Answer 1

I don't know what "dominating" means so perhaps this isn't a counterexample, but how about $-p(x)=x^5 - x^4 - x^3 - x^2 + x + 1=(x^2-x-1)(x^3-1)$ (which has a unique real root greater than 1), with $a_1,...,a_5=1,2,3,5,8$ and $s_1=s_2=-1$ and $s_3=1$ giving $\sum s_ix^i=x^3-x^2-x$?

Answer 2

One direction is true (even without $\alpha>1$).

The other directions seems to be wrong. Take the recurrence $$-a_i+a_{i-1}+a_{i-2}+a_{i-3}=0$$ with initial conditions $a_0=a_1=a_2=1$. Then $p(x)=-x^3+x^2+x+1$ has a "dominating" real root $\alpha=1.839\dots$. I am not clear about what dominating means but $\alpha$ is bigger (in magnitude) than the modulus of the other (complex) roots.

It is clear that $-a_5+a_4+a_3+a_1=0$ as well, the corresponding polynomial is $Q(x)=-x^5+x^4+x^2+1$. But, $p(x)$ does not divide $Q(x)$.

If you must insist strict monotonicity, then consider the below example. $$-a_i+a_{i-1}+a_{i-2}-a_{i-3}+a_{i-4}=0$$ with initial conditions $a_j=j$ for $j=0,1,2,3$. The sequence begins with $0,1,2,3,4,6,9,14,\dots$. Now, take $-a_7+a_6+a_5-a_1=0$. But, $P(x)=-x^4+x^3+x^2-x+1$ has a root $\alpha=1.512\dots$ and yet $P(x)$ does not divide $$Q(x)=-x^7+x^6+x^5-x=-x(x-1)(x^5-x^3-x^2-x-1).$$