Talk:3x-1 problem

Trajectories that eventually hit a power of 2

Regarding "It is very likely that the 3x-1 problem has trajectories that eventually hit a power of 2"... Shouldn't that be "there are" such trajectories? For example, the sequence 29, 86, 43, 128, 64, 32, 16, 8, 4, 2, 1, ... would be one. Or am I misunderstanding? Will Nicholes 16:44, 15 February 2011 (UTC)

You understood it correctly. I think Daniel was trying to make some other point, which I'll have to ask him about. For now, I've corrected the page to the point that you and I are understanding. By the way, a sequence containing 29, 86, 43, 128, ... as a subsequence might be worth adding to the OEIS. Alonso del Arte 17:47, 15 February 2011 (UTC)

Thanks Alonso. I've found a few interesting 3x+1 related sequences I'm planning to submit; once I'm done with that I will probably submit a couple of interesting 3x-1 related sequences as well (assuming no one beats me to them.) Will Nicholes 18:46, 15 February 2011 (UTC)

You can take your time. Right now the other contributors seem far more interested in prime numbers. Alonso del Arte 22:49, 15 February 2011 (UTC)

Daniel, I'm sure you realize that numbers of the form

${\frac {2^{2n+1}+1}{3}}\cdot 2^{k},\quad n\geq 0,\ k\geq 0,\,$

always hit 2^(2n + 1). The 2^k gets divided out in precisely k halving steps. Then the tripling step gets rid of 3 as a denominator in the fraction (or replaces it by 1 if you prefer) leaving us with 2^(2n + 1) + 1. Lastly, the subtraction of 1 gets rid of the second "+ 1," leaving precisely 2^(2n + 1).

What I don't know is if there's a pedagogical purpose to saying "seems to hit ... is this always the case?" If this was a textbook, I figure such a statement would be in the exercises section of the chapter. Alonso del Arte 06:30, 16 February 2011 (UTC)

I should have figured that one, obviously... — Daniel Forgues 08:37, 16 February 2011 (UTC)