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T. Amdeberhan
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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 $$Q(x)=-x^7+x^6+x^5-x=-x(x-1)(x^5-x^3-x^2-x-1).$$

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 $Q(x)$.

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 $$Q(x)=-x^7+x^6+x^5-x=-x(x-1)(x^5-x^3-x^2-x-1).$$

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T. Amdeberhan
  • 43.2k
  • 5
  • 57
  • 217

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 $Q(x)$.

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 $p(x)=-x^3+x^2+x+1$.

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 $Q(x)$.

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 $Q(x)$.

Source Link
T. Amdeberhan
  • 43.2k
  • 5
  • 57
  • 217

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 $p(x)=-x^3+x^2+x+1$.

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 $Q(x)$.