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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.

TMM
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gaoxinge
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1 Answers1

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

JPi
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    I'm wrong and fix it – gaoxinge Jan 26 '14 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 '14 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 '14 at 16:03
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    But it is an interesting problem! – JPi Jan 26 '14 at 16:07