Let for integers $n\geq 1$, the radical of an integer $\operatorname{rad}(n)$, see the definition from this Wikipedia. Is a famous multiplicative function since it is related to the abc conjecture. I am thinking different versions of the so-called $3x+1$ problem , that is Collatz conjecture, see this Wikipedia. There were in the literature different variations of such problem, but I don't know if next definition/variation was in the literature, or well if one can deduce easily statements in contrast with the Collatz conjecture.
Question. I define for integers $n\geq 1$ $$ f(n) := \begin{cases} \frac{\operatorname{rad}(n)}{2} & \text {if $\gcd(2,n)=0$} \\ \operatorname{rad}(3n+1) & \text {if $\gcd(2,n)=1,$} \end{cases} $$ for same recurrence of those $a_i$ of the Collatz conjecture, I am saying the recurrence of previous reference.
Q1) Do you know if my creation was in the literature? In this case refer the literature and I try to find and read some aspect of those calculations. Alternatively, provide us some feedback about if do you think that this definition should be interesting.
Q2) I know few facts about the nature of the radical of an integer, my intuition tell me that the sequences defined from this algorithm should attain an end (read a cycle containing $1$), and that should be easy to prove it. More fuzzy are my thoughts about if one can say something about the total stopping time. Is it possible to deduce some statement about the total stopping time as was defined from previous reference? Many thanks.
Example. The sequence for $n=3$ attains an end: here if there are no typos in my calculations $a_0=3$, $a_1=f(a_0)=\operatorname{rad}(3\cdot 3+1)$ since $3$ is odd, and since $10$ is squarefree one calculated $a_1$ as $10$. Thus from $a_2=f(a_1)=f(10):=\frac{10}{2}=5$ and from here one has $a_3=\operatorname{rad}(3\cdot 5+1)=2$ since $\gcd(2,5)=1$ and the only prime dividing $16$ is two. Then the calculation $a_4=\frac{2}{2}=1$ starts a cycle, because $f(1)=\operatorname{rad}(4)=2$.
One sure thing,it will alternate between $f_1(n)=rad(3n+1)$ and $f_2(n)=\frac{rad(n)}{2}$: $f_1(n)$ will produce an even number with only one factor 2,and $f_2(n)$ will cut it out. In some sense,it is equivalent to the $\frac{3n+1}{2^p}$,but at some point there will be a change of route when the $rad$ will cut other factors(e.g.3*161+1=2*2*11*11,this will be shortcuted to 11 which is not on the collatz path of 161). Perhaps it may lead to other cycles.