Talk:710: Collatz Conjecture

Technically, the 1 should lead to the four, causeing a loop. Because (1*3)+1=4, 4/2=2, 2/2=1 xD -- ‎87.242.215.66 (talk) (please sign your comments with ~~~~)

I'm pretty sure (follow the Wiki link, perhaps, although I haven't yet) there's an implicit "...eventually lead to one, then you can stop" for this process.

And (again without checking the Wiki link), I suppose the conjecture could fail in one of two cases. Firstly if multiplying by three and adding one would take us to another odd number. Which cannot happen, because (odd*3) will be odd, so (odd*3)+1 is even. Which leads us to the possibility that the even number leads back to a prior odd number, to circle around again. There's no trivial case of (n*3)+1 => m, (m/2) => n, although there is the case of (n*3)+1 => m, (m/2) => o, (o/2) => n, for n=1. How, though, could we evaluate f_1|2(x_i) => x_i+1 for all countably finite limits to i, given the rules of which f() function to use, to ensure that x_i never equals x₀. Now, that's a question and a half. Which I suspect has already been asked. And a half. 31.111.50.225 21:36, 7 May 2013 (UTC)