Let's define the following two recurrence relations with $c_{1,2} \in \mathbb{N}$ and $x \in \mathbb{R}$. The constant coefficients $c_1$ and $c_2$ are the same for both recurrence relations.
First recurrence relation (constant coefficients)
$a_0(x) = 1$
$a_1(x) = c_1 x$
$a_n(x) = c_1 x a_{n-1}(x) + c_2 x a_{n-2}(x)$ with $c_{1,2} \in \mathbb{N}$ and $x \in \mathbb{R}$
Second recurrence relation (variable coefficients)
$b_0(x) = 1$
$b_1(x) = \frac{c_1}{2} x$
$n b_n(x) = c_1 x b_{n-1}(x) + c_2 x b_{n-2}(x)$
Target of investigation
The regions of $x$ where the polynomial $b_n(x) \neq 0$.
Preparatory work
In this question (StackExchange - Mathematics) I showed that $a_n(x) \neq 0$ if the discriminant of the characteristic polynomial $\Delta_{\lambda} > 0$. Because these polynomials have only positive coefficients, all real roots lie on the negative real axis.
$\Delta_{\lambda} = x \left( c_1^2 x + 4 c_2 \right) > 0$
So $a_n(x) \neq 0$ if $x < -\frac{4 c_2}{c_1^2}$ (or $x > 0$).
Motivated from this result, I investigated into the polynomial $b_n(x)$ and got numerical evidance that $b_n(x) \neq 0$ if $x < -\frac{4 c_2}{c_1^2}n$.
Question: Is it mathematical correct if I prove this (asymptotic) behavior by the following connection with the result from $a_n(x)$ ? Otherwise: can someone show me another (better, simpler) way to prove it?
$n b_n(x) = c_1 x b_{n-1}(x) + c_2 x b_{n-2}(x)$
$b_n(x) = c_1 \frac{x}{n} b_{n-1}(x) + c_2 \frac{x}{n} b_{n-2}(x) = c_1 y b_{n-1}(x) + c_2 y b_{n-2}(x)$ with $y = \frac{x}{n}$.
So from the previous results, I know that $a_n(x) \begin{cases}> 0 & \text{ if } n \text{ is even} \\ < 0 & \text{ if } n \text{ is odd}\end{cases}$ and $x < -\frac{4 c_2}{c_1^2}$.
The relation $b_n(x)$ is similar to $a_n(x)$ such that every $x$ in $a_n(x)$ gets replaced by $\frac{x}{n}$ in $b_n(x)$. So the most reduction of $x$ would happen if I set $n$ in the recurrence relation to the current value of $n$.
For example, I take a new recurrence relation instead of $b_5(x)$ I use:
$b_0^{(*)}(x) = 1$
$b_1^{(*)}(x) = \frac{c_1}{5} x$
$b_n^{(*)}(x) = c_1 \frac{x}{5} b_{n-1}^{(*)}(x) + c_2 \frac{x}{5} b_{n-2}^{(*)}(x)$
I then can use again the results from the investigation for constant coefficients which $b_n^{(*)}(x) \neq 0$ if $\frac{x}{5} < -\frac{4 c_2}{c_1^2}$.
And for every $b_n(x)$ I would define a new recurrence relation such that at the end I am able to say that:
$b_n(x) \neq 0$ if $\frac{x}{n} < -\frac{4 c_2}{c_1^2}$ or respectively $x < -\frac{4 c_2}{c_1^2} n$