September 2015

[LaurV]

Like one of my friends would say, are you arguing with me, or against me?
You said exactly what I said, with different words, and you may have also found "the little trick" I was talking at the end of my post.

OTOH, YES, your solutions, all 4, are "best", i.e. they can not be improved.
For example for n=12, the N=1033 is optimal, because the only improvement you could do (from the "2^x+1 rule") would be to raise the 256 to 257. In this case is easy to see that no matter what you use for the next two values, x, y, there is no way to reward 9 teams.

2,3,5,9,17,33,65,129,256257,261x,772y=1034-x,10331034 - No way to reward 9 teams.

[R. Gerbicz]

Originally Posted by LaurV

OTOH, YES, your solutions, all 4, are "best", i.e. they can not be improved.
For example for n=12, the N=1033 is optimal, because the only improvement you could do (from the "2^x+1 rule") would be to raise the 256 to 257. In this case is easy to see that no matter what you use for the next two values, x, y, there is no way to reward 9 teams.

Here some proof(s) could be better, not just numerology. And actually my program *proved* the optimality for n=9,10, not just found "good" solutions.