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Apr 12 at 2:53 answer added bof timeline score: 9
Apr 12 at 2:53 comment added Daniel Weber The set of squares also works
Apr 12 at 2:37 review Close votes
Apr 18 at 3:02
Apr 12 at 2:29 comment added Nick S Not a direct answer to this problem, but the square free integers $S$ have the property that $$ \mathrm{Card}(S \pmod p^2)=p^2-1$$ for all primes $p$. Moreover, if I am not mistaken, a set $X$ belongs to the dynamical system generated by $S$ exactly when $$\mathrm{Card}(S \pmod p^2)<p^2$$ for all primes, and I find this last bit very interesting.
Apr 12 at 2:27 history edited joaopa CC BY-SA 4.0
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Apr 12 at 2:14 comment added mathworker21 $S := \{(2n)! : n \in \mathbb{N}\}$
Apr 12 at 2:02 history asked joaopa CC BY-SA 4.0