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Feb 19, 2019 at 9:13 answer added Oliver Roche-Newton timeline score: 7
Feb 15, 2019 at 16:18 comment added LeechLattice Surely, the sequence works as an upper bound.
Feb 15, 2019 at 15:56 comment added Greg Martin @Bullet51: Note that A066063 is only a related sequence, not the same one, since the OP is concerned with modular arithmetic.
Feb 15, 2019 at 13:53 comment added LeechLattice oeis.org/A066063
Feb 15, 2019 at 10:59 comment added Fedor Petrov Obviously $m(m+1)/2\geqslant N$ where $m=|X|$. On the other hand, therу exist such $X$ of size roughly $C\sqrt{N}$ (take $s=[\sqrt{N}]$ and $X=\{0,1,\dots,s-1,s,2s,3s,\dots,s^2,s(s+1)\}$.)
Feb 15, 2019 at 10:56 history asked ililiil CC BY-SA 4.0