This question is an extension of Why are the roots of this recursive defined polynomial bound by the roots of the discriminant of the characteristic polynomial?
As I discovered numerically, I can extend this recurrence polynomial to:
$$ \begin{align} p_0(x) &= 1 \\ p_n(x) &= \frac{x}{f(n)} \sum\limits_{k=1}^n a_k p_{n-k}(x) \end{align} $$
where $a_n$ is an arbitrary arithmetic progression like $a_n = a_1 + (n-1) d$ and $f(n)$ is a monotonic increasing function with positive values like $f(n) = n$, $f(n) = n^2$ or even $f(n) = \ln(n+1)$.
Then for the roots $x_0$ of the polynom $p_n(x_0) = 0$ the following inequality holds:
$x_0 > - \frac{4d}{a_1^2} f(n)$
It seems to be such an easy rule, but I don't know if it is easy to prove such like defining $y = \frac{x}{f(n)}$ and using $p_n(y)$. There seems to be an asymptotic behavior.
Here some examples:
Let be $a_1 = 1$ and $d = 1$ such that $a_n = n$ then all the roots of the polynomials $p_n$ are located inside the region defined by $-4 f(n) < x_0 \leq 0$.