22 Jun
2018
22 Jun
'18
9:23 p.m.
It turned out to be quite easy to prove once I looked at it the right way. It’s equivalent to the fact that n(n+1)/2 is 0 mod n iff n is odd. Jim Propp On Friday, June 22, 2018, James Propp <jamespropp@gmail.com> wrote:
On Friday, June 22, 2018, Marc LeBrun <mlb@well.com> wrote:
It appears to have the alternating form 1 0, 2 1's, 3 0's, 4 1's, 5 0's
and so on. Are you asking if the diagonal selection process from the array has this apparent structure in fact?
Yes.
Jim Propp