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$n$,$k$, $m$, $u$ $\in$ $\Bbb N$;

Let's see the following sequence:

$x_0=n$; $x_m=3x_{m-1}+1$.

I am afraid I am a complete noob, but I cannot (dis)prove that the following implies the Collatz conjecture:

$\forall n\exists k,u:x_k=2^u$

Could you help me in this problem? Also, please do not (dis)prove the statement, just (dis)prove it is stronger than the Collatz conjecture.

If it implies and it is true, then LOL.


Okay, let me reconfigure the question: let's consider my statement true. In this case, does it imply the Collatz conjecture?

Please help me properly tagging this question, then delete this line.

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@AndréNicolas: OK. How do you know? – Did Jul 27 '12 at 10:04
@did: Misread.${}{}$ – André Nicolas Jul 27 '12 at 10:19
About your update, let me repeat: let us consider your statement true; then the integer $6$ is odd, the number $\pi$ is an integer, Collatz conjecture holds and Collatz conjecture does not hold. – Did Jul 31 '12 at 7:48

Call your statement S. Then: (1.) S does not hold. (2.) S implies the Collatz conjecture (and S also implies any conjecture you like). (3.) I fail to see how Collatz conjecture should imply S. (4.) If indeed Collatz conjecture implies S, then Collatz conjecture does not hold (and this will make the headlines...).

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I see from a false statement everything is implied, but I think in this exact situation we needn't investigate the truthness of $S$. – vakufo Jul 27 '12 at 10:13
In principle you may be right. In the case at hand however, (1.) and (4.) destroy any hope that a simple proof of the equivalence of S and Collatz conjecture exists . – Did Jul 27 '12 at 10:17
From an integer $n$ of the shape $4t+3$ we can never reach a power of $2$. – André Nicolas Jul 27 '12 at 10:34

Even if we make believe that we don't know that the statement is false, it is still not clear whether it says anything about Collatz. Collatz doesn't go by doing $3x+1$ until you get a power of 2; it goes by alternating doing $3x+1$ and dividing out powers of 2. Every time you divide out a power of 2, you get to a new odd number, and have to start the $3x+1$ bit all over again. There's no reason to think you'd ever hit a power of 2.

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Indeed (my point (3.)). – Did Jul 27 '12 at 17:52

I think your statement is not true, and even aside from its falseness, it simply does not imply Collatz.

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