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A077567
Least k >= 2 such that sigma(n) divides sigma(n^k).
1
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, 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, 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
OFFSET
1,1
COMMENTS
From Robert Israel, Mar 14 2017: (Start)
If x and y are coprime and a(x)=a(y)=k, then a(xy)=k as well.
If n > 1 is squarefree, then a(n^k) = k+2 for all k>=1.
Is there any n > 1 with a(n) = 2? (End)
LINKS
MAPLE
f:= proc(n) local k, s; uses numtheory;
s:= sigma(n);
for k from 2 do if sigma(n^k) mod s = 0 then return k fi
od
end proc:
map(f, [$1..200]); # Robert Israel, Mar 14 2017
MATHEMATICA
a[n_] := For[k = 2, True, k++, If[Divisible[DivisorSigma[1, n^k], DivisorSigma[1, n]], Return[k]]];
Array[a, 100] (* Jean-François Alcover, Aug 24 2020 *)
CROSSREFS
Sequence in context: A139069 A071866 A077603 * A096344 A030349 A285203
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Dec 01 2002
STATUS
approved