See this
After seeing this question, I observed first 10 natural numbers , I saw this
For every $n\in \mathbb N$ and $n\ne 2^k$ for some $k\in \mathbb N$ , after applying these two operations , these terms must occurs, $$10,5,16,8,4,2$$ Only thing I want is a counter example or a satisfoctary answer about the truth of this fact
Note that in case of $5, 10$ will be ommitted from the list , but it follows for every other $n$
I am still checking for more numbers to found a counter example myself
But since I am doing it with a copy pen, I can never reach a conclusion
Please help!!!
Starting from $21$ gives you $$21,64,32,16,8,4,2,1.$$ Starting from $85$ gives you $$85,256,128,64,32,16,8,4,1.$$ In general, every power of $4$ is of the form $3m+1$, with $m$ necessarily odd; so you can always find odd numbers that get mapped to a power of $4$, and thus the induced sequence does not pass contain $10$ nor $5$.