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A077567
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Least k >= 2 such that sigma(n) divides sigma(n^k).
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1
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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
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OFFSET
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1,1
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COMMENTS
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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)
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LINKS
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MAPLE
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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:
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MATHEMATICA
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a[n_] := For[k = 2, True, k++, If[Divisible[DivisorSigma[1, n^k], DivisorSigma[1, n]], Return[k]]];
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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