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Comment by remus | Hacker News Reader

Depends what you mean by a good reason I guess. A lot of very capable mathematicians have put time in to the problem and it's proven to be pretty resistant to a solution. I think problems like this often feel a little incongruous because they are so simple to state but so hard to solve.

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It is easy, we can spit it into two parts: Proving that all even numbers end up in that loop and proving that all odd numbers do.

I'll take the even ones. Who works on the odd ones?

I'll take a stab.

An odd number is a number that ends in 1, 3, 5, 7, or 9.

- ends in a 1: times 3 + 1: it now ends in a 4, which is even, and thus reduces to the even case.

- ends in a 3: tomes 3 is 9, + 1: ends in a 0; even again.

- ends in a 5: ×3 ends in a 5 again, +1: ends in a 6; again even!

- ends in 7: x3: now ends in a 1, +1: ends in a 2; even yet again!

- ends in a 9: ×3 then eens in a 7, +1 thus eens in an 8: even again!!

So all that's left is the cases for @someweirdperson ;-) @someweirdperson

I think this is getting at a pretty good point. Half of all numbers (all evens) will scale by .5 to the subsequent number. The other half (odds) will effectively scale by approximately 3/2, as you have shown here. Obviously over a large amount of iterations this should cause the calculations to trend lower and lower.

One difficulty with this approach is that it doesn't help disproving the existence of cycles, it only makes it less likely that the sequence diverges to infinity.

Another approach would be using the "inverse" of the formula and then trying to prove that you can build all natural numbers starting from 1.

This approach, like every other approaches, did not yield any significant result.

The funny thing is that it fits right in the P=NP problem: easy to verify, hard to solve :)

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