Explain xkcd: It's 'cause you're dumb.
Technically, the 1 should lead to the four, causeing a loop. Because (1*3)+1=4, 4/2=2, 2/2=1 xD 188.8.131.52 (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 f1|2(xi) => xi+1 for all countably finite limits to i, given the rules of which f() function to use, to ensure that xi never equals x0. Now, that's a question and a half. Which I suspect has already been asked. And a half. 184.108.40.206 21:36, 7 May 2013 (UTC)
- It seems way too general to be much more than "asked," and I am sure that it has been addressed in its simpler forms. In any case, there is enough amateur, recreational, and serious mathematical literature on it to find out that there are indeed two failure cases: a starting Collatz number results in an infinitely increasing sequence, or a loop exists apart from the 4-2-1 loop. (Curiously enough, some loops exist when negative numbers are allowed.) Stuff like this and Goldbach made me realize just how hard simple things can get. --Quicksilver (talk) 03:04, 20 August 2013 (UTC)