Collatz finally solved? [closed]

http://www.newscientist.com/blogs/shortsharpscience/2011/06/simple-number-puzzle-possibly.html

What does mathematicians say of this proof, right or wrong?

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closed as off topic by Aryabhata, Jonas Meyer, Byron Schmuland, Chris Eagle, Eric NaslundJun 3 '11 at 21:15

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Probably no one knows yet. In general the track record for such announcements has been pretty poor.. but hey you never know :) –  Zarrax Jun 3 '11 at 16:56
@ MartianInvader: Why? Any operator of the form $f \mapsto f \circ g$ is linear. –  Mark Jun 3 '11 at 17:52
Looks like "not a real question" to me (I mean, it has definitive answer, and it will be given — but it will be given by professional community; and using math.SE question just for collecting links is clearly wrong) –  Grigory M Jun 3 '11 at 17:55
@TonyK To make the long story short, SE is no Reddit. But if you're surprised by the very idea of following SE rules, I don't know what to say. –  Grigory M Jun 3 '11 at 18:26
Related meta thread –  Grigory M Jun 3 '11 at 18:49

Here's a short summary of "what it all boils down to".

Let $k$ be some arbitrary integer. Define a sequence by iterating the mapping $k \mapsto (3k+1)/2$ until you hit an even number. For example, the sequence for $7$ is $$7, 11, 17, 26.$$ Denote the even number which terminates the sequence by $F(k)$. So $F(7) = 26$.

We now define a directed graph on all integers $\geq 3$. An edge $j \rightarrow j"$ exists if the sequence for $j"$ contains $F(j)/2$.

For example, we always have $2j \rightarrow j$ and $j \rightarrow F(j)/2$.

Opfer reduces the Collatz conjecture to showing that the vertex $4$ is reachable from any vertex $n \geq 3$. The proof of this fact can be found on page 11.

Note: I obtained this summary by "diagram chasing" across the paper. It's not presented in exactly that way.

Update: On June 17th, 2011 Opfer's pre-print available here has been updated with this comment:

Author's note: The reasoning on p. 11, that "The set of all vertices (2n; l) in all levels will contain all even numbers 2n 6 exactly once." has turned out to be incomplete. Thus, the statement that "the Collatz conjecture is true" has to be withdrawn, at least temporarily.

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Might I add that "this fact" is plainly the same as the Collatz conjecture, and the very sketchy argument on page 11 is not very convincing. –  Yuval Filmus Jun 3 '11 at 19:54
Truth - not as clean as I had hoped from reading the first page. –  mixedmath Jun 3 '11 at 21:27
Yes, so the re-interpretation of the problem in terms of the kernel of some linear operators doesn't seem to come into it when you start to dig down. And the sentence "The set of all vertices $(2n,l)$ in all levels will contain all even numbers $2n\ge6$ exactly once." on page 11 is a restatement of the Collatz conjecture, but I haven't worked out what this statement follows from. –  George Lowther Jun 3 '11 at 23:16
Here is a pointer to some other people who've come to the same conclusion: mathlesstraveled.com/2011/06/04/…. –  Yuval Filmus Jun 4 '11 at 23:59
The preprint now contains an "Author's note" at the front that says "The reasoning on p. 11, that 'The set of all vertices $(2n,l)$ in all levels will contain all even numbers $2n \ge 6$ exactly once.' has turned out to be incomplete. Thus, the statement 'that the Collatz conjecture is true' has to be withdrawn, at least temporarily". June 17, 2011 –  joriki Jul 20 '11 at 7:28

I don't know whether the paper is right, but it sure looks like a legitimate math paper, and seems to represent the culmination of decades of work by various mathematicians. The author is a reputable mathematician at a major university, and the math genealogy page indicates that he was originally a student of Lothar Collatz.

Apparently the preprint has been submitted to Mathematics of Computation, so it will now undergo the formal peer review process. It is much too early to tell how this will turn out.

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protected by Qiaochu YuanJun 3 '11 at 17:07

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