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A254733 a(n) is the least k > n such that n divides k^3.

%I A254733 #27 Nov 23 2024 12:34:11
%S A254733 2,4,6,6,10,12,14,10,12,20,22,18,26,28,30,20,34,24,38,30,42,44,46,30,
%T A254733 30,52,30,42,58,60,62,36,66,68,70,42,74,76,78,50,82,84,86,66,60,92,94,
%U A254733 60,56,60,102,78,106,60,110,70,114,116,118,90,122,124,84
%N A254733 a(n) is the least k > n such that n divides k^3.
%C A254733 A073353(n) <= a(n) <= 2*n. Any prime that divides n must also divide a(n), and because n divides (2*n)^3.
%H A254733 Peter Kagey, <a href="/A254733/b254733.txt">Table of n, a(n) for n = 1..5000</a>
%F A254733 a(n) = n + A019555(n).
%e A254733 a(8) = 10 because 8 divides 10^3, but 8 does not divide 9^3.
%t A254733 lkn[n_]:=Module[{k=n+1},While[PowerMod[k,3,n]!=0,k++];k]; Array[lkn,70] (* _Harvey P. Dale_, Nov 23 2024 *)
%o A254733 (Ruby)
%o A254733 def a(n)
%o A254733   (n+1..2*n).find { |k| k**3 % n == 0 }
%o A254733 end
%o A254733 (PARI) a(n)=for(k=n+1,2*n,if(k^3%n==0,return(k)))
%o A254733 vector(100,n,a(n)) \\ _Derek Orr_, Feb 07 2015
%Y A254733 Cf. A073353 (similar, with k^n).
%Y A254733 Cf. A254732 (similar, with k^2), A254734 (similar, with k^4).
%Y A254733 Cf. A019555 (similar without the restriction that a(n) > n).
%K A254733 nonn,easy
%O A254733 1,1
%A A254733 _Peter Kagey_, Feb 06 2015