Collatz conjecture

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

• @Arnaud.D I want a counter example if this is false or an explanation why this is true – Atul Mishra Jun 27 '17 at 15:25
• Not sure why this got two downvotes - it's a reasonable question to ask, I think. – Noah Schweber Jun 27 '17 at 15:38

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 $5$.

• +1. For the OP, a proof of the statement that every power of $4$ is of the form $3m+1$ with $m$ odd: to see that every power of $4$ is of the form $3m+1$, note that $4$ itself is, and $(3m+1)(3n+1)=9mn+3n+3m+1=3(3mn+m+n)+1$ is of the form $3k+1$ (more concisely: $4=1 (mod 3)$, so $4^k=1^k(mod 3)=1(mod 3)$). And if $m$ is even, then $3m+1$ is odd, so if $4^k=3m+1$ then $m$ must be odd. – Noah Schweber Jun 27 '17 at 15:38