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Nov 6, 2019 at 17:24 vote accept Mishel Skenderi
Nov 6, 2019 at 5:03 comment added Noam D. Elkies Also condition (2) is automatic from (1) once $S \neq \emptyset$: if $S \ni t_1$ then $S$ also contains $2t_1, 3t_1, 4t_1, \ldots$, so there is never a gap longer than $t_1$.
Nov 6, 2019 at 5:01 answer added Noam D. Elkies timeline score: 6
Nov 6, 2019 at 1:52 comment added Mishel Skenderi Thank you; in fact, in my particular context, I also always assume $n \geq 2.$
Nov 6, 2019 at 1:50 history edited Mishel Skenderi CC BY-SA 4.0
edited body
Nov 6, 2019 at 1:41 comment added Noam D. Elkies Probably you want to require $n \geq 2$, because for $n=1$ the only allowed $S$ are arithmetic progressions $-$ but there are non-AP's that satisfy (1,2,3), such as the positive integers not congruent to $1$ or $5 \bmod 6$.
Nov 6, 2019 at 1:03 history edited Mishel Skenderi CC BY-SA 4.0
fixed typo
Nov 6, 2019 at 0:55 history edited Mishel Skenderi CC BY-SA 4.0
added condition 3)
Nov 6, 2019 at 0:05 review First posts
Nov 6, 2019 at 5:37
Nov 6, 2019 at 0:02 history asked Mishel Skenderi CC BY-SA 4.0