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%I #17 Aug 24 2020 10:44:43
%S 2,3,3,4,3,3,3,5,4,3,3,5,3,3,3,6,3,7,3,5,3,3,3,3,4,3,5,7,3,3,3,7,3,3,
%T 3,4,3,3,3,5,3,3,3,5,3,3,3,11,4,7,3,7,3,5,3,3,3,3,3,5,3,3,7,8,3,3,3,5,
%U 3,3,3,13,3,3,7,5,3,3,3,5,6,3,3,5,3,3,3,5,3,3,3,5,3,3,3,7,3,4,7,4,3,3,3
%N Least k >= 2 such that sigma(n) divides sigma(n^k).
%C From _Robert Israel_, Mar 14 2017: (Start)
%C If x and y are coprime and a(x)=a(y)=k, then a(xy)=k as well.
%C If n > 1 is squarefree, then a(n^k) = k+2 for all k>=1.
%C Is there any n > 1 with a(n) = 2? (End)
%H Robert Israel, <a href="/A077567/b077567.txt">Table of n, a(n) for n = 1..10000</a>
%p f:= proc(n) local k, s; uses numtheory;
%p s:= sigma(n);
%p for k from 2 do if sigma(n^k) mod s = 0 then return k fi
%p od
%p end proc:
%p map(f, [$1..200]); # _Robert Israel_, Mar 14 2017
%t a[n_] := For[k = 2, True, k++, If[Divisible[DivisorSigma[1, n^k], DivisorSigma[1, n]], Return[k]]];
%t Array[a, 100] (* _Jean-François Alcover_, Aug 24 2020 *)
%K nonn
%O 1,1
%A _Benoit Cloitre_, Dec 01 2002
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