wilmott.com

The Collatz conjecture has been discussed at least once here, and I was wondering if one needs to prove it only for the even integers, since if the original chosen number is odd, the first operation always results in an even number. So if one can prove the conjecture for all even numbers, wouldn't that cover (after one "step") all the odd numbers as well?Thanks for any thoughts!MikeCollatz conjecture: Link

Certainly, as the algorithm would terminate (at 1) for an odd n if it does so for 2n...