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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, Daniel Rust, Hagen von Eitzen, hardmath, Thomas Andrews Jan 26 at 13:24

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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 at 12:13
    
@TMM Open problem? –  gaoxinge Jan 26 at 12:21
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Thank you @DanielRust –  gaoxinge Jan 26 at 12:31
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Why downvote? It's a legitimate question and the OP never heard of this Collatz conjecture before. –  Alex Jan 26 at 13:40

1 Answer 1

up vote -1 down vote accepted

It's false because if $a_n=1$ then $a_{n+1}=4$.

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I'm wrong and fix it –  gaoxinge Jan 26 at 11:52
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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 at 13:42
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Well, I tried in earnest to answer the OP's question and I understand that it's easy to make mistakes; we all do it all the time. But I'm not telepathic and cannot guess that the OP actually did make a mistake given that I didn't know the Collatz conjecture. So I'm answering the question as stated. –  JPi Jan 26 at 16:03
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But it is an interesting problem! –  JPi Jan 26 at 16:07

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