If $f(x)=3x+2^n$ where $2^n$ factors $x$, is $f(x)$ coprime with $x/2^m$?

Question

If $O$ is the positive odd integers and $X$ the positive even integers then for every $x_n\in X$ we can define the product of its odd factors $o_n\in O$ and it is self-evidently true that $X=\{x_n:x_n=o_n\times 2^m:m\in\mathbb{N}, m>0\}$

Let $$x_{n+1}=f(x_n)=3x_n+2^m$$

It is plain that $o_{n+1}$ is coprime with $o_n$

Is it also true that $o_{n+2}$ is coprime with $o_n$?

Then by induction, $o_n$ would be coprime with every $o_p:p<n$, would it not? Else can we show this by other means?

Answer 1

This isn't true, and here is a counterexample: consider $x_1=30=2\cdot 15$. Then $x_2=3\cdot 30+2=92=2^2\cdot 23$, and $x_3=3\cdot 92+2^2=280=2^3\cdot 35$, so $o_3$ and $o_1$ are both divisible by $5$.