John

Well, all odd natural numbers, though I've been a bit sloppy myself, simply saying "odd numbers." For integers in general, there are at least four non-trivial cycles in addition to the trivial cycle 0 -> 0.

Certainly. And I guess the hope (and what you'd like to prove) is that after some N, all odd numbers will be accounted for, so the iteration will cease and we'll know that the Collatz conjecture holds.

So you iterate through the natural numbers, and for each natural number N, the equation takes that and outputs all odd numbers whose sequences contain exactly N iterations of the odd rule, along with the powers of 2 involved in traveling from one odd number in the sequence to another?

So for N = 1, we have {1, 5, 21, 85, 341...}, or [(4^n) - 1]/3 for all natural numbers n. I'm eager to see how the equation produces/presents its output.

Yeah, it seemed like the kind of thing you'd have realized fairly quickly, but I decided to simply edit rather than delete just in case. After all, it's easy to miss the trees for the forest sometimes.

So just so I'm clear here, you're developing a formula that, given an odd input c, should generate all the numbers that reduce to 1 using n/2 for evens and cn+1 for odds along the sequence? Or what?

The 1n+1 case is trivially true, since (n+1)/2 < n except for the case n = 1. So you can probably take that as given and move on to looking at the relationships between various coefficients.

Edit: Unless you mean you're trying to prove it using the equation you're developing, in which case don't mind me.

The forums were created in early July of 2002, so call it a little under 11.5 years.

I actually need to some time soon. I have people around the world to captivate with the beauty of my handwriting.

Makes me feel so good, makes me feel so numb, yeah.

Fabric of the Cosmos deals more with general modern physics than with string theory in particular. As such, I found I enjoyed it more than I did The Elegant Universe.

It's a tongue-twister my dad taught me when I was very young. I guess it's a British thing. In any case, he could say it extremely quickly. I think the slur inherent to my southern American accent hinders my efforts somewhat.