Let $a_1$ be an integer. Then we assume

$$a_{n+1} = \begin{cases} 3a_n+1,&\text{a_n is odd}\\ \frac{a_n}{2},&\text{a_n is even} \end{cases}$$

Now we prove that

for any $a_1\in\mathbb N$, there exists $N$ which satisfy: $a_n=1,2$ or $4$,$n\geq{N}$.

At first I want to give it a suitable category for the problem: analysis. And I want to use the basic method: evaluate the upper bound for $a_n$, however I find it's not easy because the iteration is rely on the odd or even property of $a_n$. So I attempt the method of number theory. But I failed to find any way to go over it. Can anyone have idea? Thank you.

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marked as duplicate by TMM, Dan Rust, Hagen von Eitzen, hardmath, Thomas AndrewsJan 26 '14 at 13:24

This question was marked as an exact duplicate of an existing question.

This is a long-standing open problem from number theory, which you are not going to solve with "basic methods". Voting to close. – TMM Jan 26 '14 at 12:13
@TMM Open problem? – gaoxinge Jan 26 '14 at 12:21
Thank you @DanielRust – gaoxinge Jan 26 '14 at 12:31
Why downvote? It's a legitimate question and the OP never heard of this Collatz conjecture before. – Alex Jan 26 '14 at 13:40

It's false because if $a_n=1$ then $a_{n+1}=4$.
I realize you answered an early statement of the problem, but it really wasn't so badly stated that we could not interpret it correctly (as the Collatz conjecture), a matter of how $n \ge N$ should be quantified. If you are posting a one-line Answer, please give the Question the benefit of any doubt. – hardmath Jan 26 '14 at 13:42