Consider a modified version of Collatz sequence:
$C(n)=\left\{ \begin{array}{ll} \frac{3n+1}{2} & n\ \mathrm{odd} \\ \frac{n}{2}& n\ \mathrm{even}\end{array} \right.$
Let $F_n$ be the smallest integer that satisfies $C^{F_n}(n)=1$, where $C^0(n)=n$, $C^1(n)=C(n)$, and $C^k(n) = C(C^{k-1}(n))$ for any integer $k>1$.
Now for any integer $n>0$ with a finite $F_n$ (can be infinite if Collatz conjecture is disproved) consider polynomial
$T_n(z)=n z^{F_n} + C(n) z^{F_n-1} + C^2(n) z^{F_n-2} + \cdots + C^{F_n-1} z + 1 = \sum_{k=0}^{F_n} C^k(n) z^{F_n-k}$
with roots $r^{(n)}_1, r^{(n)}_2, \ldots, r^{(n)}_{F_n}$.
Question: Can one prove that for any $n\neq9$, all roots are contained in an open disk of radius 1.5 around $(0,0)$, i.e.
$\forall 1\leq k \leq F_n, |r^{(n)}_k|<1.5$.
Note that $T_9(z)$ has a root at $r \approx -1.522093720599496 $. I have checked this for all $n<300000$, which obviously is not a large number, but it's striking to see how close some of the roots can get to the boundary of the disk but none pass it.