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If $O$ is the positive odd integers and $X$ the positive even integers then for every $x_n\in X$ we can define the product of its odd factors $o_n\in O$ and it is self-evidently true that $X=\{x_n:x_n=o_n\times 2^m:m\in\mathbb{N}, m>0\}$

Let $$x_{n+1}=f(x_n)=3x_n+2^m$$

It is plain that $o_{n+1}$ is coprime with $o_n$

Is it also true that $o_{n+2}$ is coprime with $o_n$?

Then by induction, $o_n$ would be coprime with every $o_p:p

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    Your notation is a bit ambiguous. If you say that $x_{n+1}:=3x_n+2^{o_n}$, then all $x_{i}$ are defined as soon as you know what $x_1$ is. So what is $x_1$? is it $2$? – 2017-01-28
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    @Mathematician42 $x_1$ is any starting even integer. This is equivalent to there being no non-trivial loop in the Collatz conjecture. – 2017-01-28
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    It's a sequence $x_1, x_2, x_3$. You will see that although the x's don't follow the Collatz trajectory, the o's do. – 2017-01-28
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    I have no idea whether $o_{n+2}$ is coprime to $o_n$. I couldn't find a counterexample (even with a computer). I guess it's true but I don't know how to show that. – 2017-01-28
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    Yeah, I'm not going to think about this, I tried a computer and failed, I have absolutely no clue how to prove or disprove such things, my lifetime is finite and I don't feel like this is the kind of problem I can tackle. Not very heroic of me, but I give up before even starting. – 2017-01-28

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This isn't true, and here is a counterexample: consider $x_1=30=2\cdot 15$. Then $x_2=3\cdot 30+2=92=2^2\cdot 23$, and $x_3=3\cdot 92+2^2=280=2^3\cdot 35$, so $o_3$ and $o_1$ are both divisible by $5$.

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    Hi Wojowo. Thanks for having a look. In your example $x_2=3\times52+2^2$. At least that's what I intended! Let me check my notation. – 2017-01-28
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    @RobertFrost This is definitely not what you wrote. Writing $x_n=2^mo_n$ means that for $x_n=52$ we have $o_n=13$. – 2017-01-28
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    Haha! Let me fix. – 2017-01-28
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    Done. Sorry for wasting your time :( It was right in the title. – 2017-01-28
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    @RobertFrost Edited. – 2017-01-28
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    yes I agree, if you just make clear the answer is no, I can accept your answer. But you should have a look at this one: http://math.stackexchange.com/questions/2117454/is-there-no-non-trivial-cycle-on-the-collatz-conejcture – 2017-01-28
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    @RobertFrost Made that clear. I have seen that other question; to me it simply seems like reducing a problem to a more complicated one, and it doesn't seem like it's in any way more tractable. – 2017-01-28
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    I have perhaps written it out in a complicated way. It simply says that $\{4,4,4,...\}$ is the only sequence of $2^{n_k}$ satisfying the main equation half-way down. So it could be proven by supposing any element of $N_K$ is not $2$ – 2017-01-28
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    I did understand that. This still doesn't make it seem any simpler to show, mainly because of the complexity of the equation. – 2017-01-28
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52664/discussion-between-robert-frost-and-wojowu). – 2017-01-28