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S Apr 26, 2016 at 1:33 history bounty ended CommunityBot
S Apr 26, 2016 at 1:33 history notice removed CommunityBot
S Apr 18, 2016 at 0:06 history bounty started Michał Masny
S Apr 18, 2016 at 0:06 history notice added Michał Masny Draw attention
Apr 23, 2015 at 0:40 vote accept Michał Masny
Apr 17, 2015 at 19:28 comment added Michał Masny @Mirko I think it's simpler to think about the case of sets containing zero only.
Apr 16, 2015 at 11:29 answer added J.-E. Pin timeline score: 6
Apr 16, 2015 at 7:31 comment added Mirko my comment above assumes that $\langle A\rangle=\mathbb N$. I may not understand what is "the semigroup generated by $A$". If $A$ contains all positive multiples of $3$, and all multiples of $5$ starting with $15$, is then $\langle A\rangle = \{3,6,9,12,15,18,20,21,23,24,25,...\}$ thus containing all positive integers beyond $23$ (yet not containing $1,2,4,5,7,8,10,11,13,14,16,17,19,22$)? Does it have to contain $0$? I guess $A$ itself (and hence $\langle A\rangle$ too) should contain $0$ (since you require it explicitly).
Apr 16, 2015 at 6:19 comment added Mirko Fix $n$ and try to see which $A$ work for this $n$. For example, for $n=3$, the set $A=\{0,1,4,7,10,...,3k+1,...\}$ easily works, but so also does the set $A=\{0,1,3,7,12,18,..\}$ even though for the latter the gaps between consecutive numbers become bigger and bigger (and even though I don't know what is the common term). We could have used $23$ instead of $18$, but making bigger gaps early may necessitate making smaller gaps later. I am pretty sure that for $n=3$ the set of triangular numbers $A=\{0,1,3,6,10,15,21,..,\dfrac{n(n+1)}2,..\}$ works(?) providing example where the gaps get larger.
Apr 15, 2015 at 21:16 history asked Michał Masny CC BY-SA 3.0