I had something similar in mind: starting with 1, can you generate all the naturals using just n' = 2n and n' = (2n+1)/3 ?Absane said:Mostly I try to work on the problem in reverse.. that is, given you start with an 2<sup>x</sup>, you can produce all the odd numbers.

I'll admit I haven't thought about this all that much, though. I like math, but it's not the main focus of my studies.

Cool. I don't know much about primes, though. Good luck with your conjecture (and thanks for reminding me of a way to spend (waste?) my summer...).Edit: I suspect this problem has a deep connection with the distibution of the prime numbers.

By the way, there's a way less useful $100 problem I've also tried (I think it was proposed by Martin Gardner): Can you find a 3x3 magic square consisting of only square numbers? I assume in the original statement of the problem, no number could appear in the square more than once and 0 wasn't allowed.