Difference between revisions of "Talk:710: Collatz Conjecture"
(Sign your posts, dammit.) |
|||
| Line 1: | Line 1: | ||
Technically, the 1 should lead to the four, causeing a loop. Because (1*3)+1=4, 4/2=2, 2/2=1 xD {{unsigned|87.242.215.66}} | Technically, the 1 should lead to the four, causeing a loop. Because (1*3)+1=4, 4/2=2, 2/2=1 xD {{unsigned|87.242.215.66}} | ||
| + | :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<sub>1|2</sub>(x<sub>i</sub>) => x<sub>i+1</sub> for all countably finite limits to i, given the rules of which f() function to use, to ensure that x<sub>i</sub> never equals x<sub>0</sub>. Now, ''that's'' a question and a half. Which I suspect has already been asked. And a half. [[Special:Contributions/31.111.50.225|31.111.50.225]] 21:36, 7 May 2013 (UTC) | ||
Revision as of 21:36, 7 May 2013
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 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. 31.111.50.225 21:36, 7 May 2013 (UTC)
