20 Oct 12, 11:11PM
(This post was last modified: 20 Oct 12, 11:50PM by Roflcopter.)
So Enam here provides a proof that all numbers divisible by an odd prime must be writeable in this way.
For odd numbers 2n+1 = n + (n+1).
For even numbers n with an odd prime factor p, n/p is an integer which we can centre a sequence of consecutive numbers around. That is n = sum(j+i, i=0..p-1) with j = n/p - (p-1)/2.
But it remains to prove that 2^m cannot be written in this way (for m>0).
For odd numbers 2n+1 = n + (n+1).
For even numbers n with an odd prime factor p, n/p is an integer which we can centre a sequence of consecutive numbers around. That is n = sum(j+i, i=0..p-1) with j = n/p - (p-1)/2.
But it remains to prove that 2^m cannot be written in this way (for m>0).