A066831 Numbers k such that sigma(k) divides sigma(phi(k)).

1, 13, 71, 87, 89, 181, 203, 305, 319, 362, 667, 899, 1257, 1363, 1421, 1525, 1711, 1798, 1889, 2407, 2501, 2933, 3103, 4609, 4615, 4687, 4843, 5002, 5191, 6583, 7123, 7625, 7627, 9374, 9947, 10063, 10411, 10991, 11107, 12989, 13543, 13891, 14587

Offset: 1

Author

Benoit Cloitre, Jan 19 2002

Keywords

• nonn

Comments

For odd n, if sigma(phi(n))/sigma(n)=3 then sigma(phi(2*n))/sigma(2*n)=1. - Vladeta Jovovic, Jan 21 2002.

Comments from Vim Wenders, Nov 01 2006: (Start)

This is almost certainly false for even n. For odd n we have phi(n)=phi(2n) and with sigma(2)=3 trivially sigma(phi(n))/sigma(n)=3 <=> sigma(phi(2n))/sigma(2n) = sigma(phi(n))/3.sigma(n)=1.

But suppose n=2m, m odd: again with phi(2m)=phi(m) and sigma(2)=3, sigma(phi(2m)) / sigma(2m)=3 => sigma(phi( m)) /3sigma( m)=3 => sigma(phi( m)) / sigma( m)=9; and with sigma(4)=7 sigma( phi(4m))/ sigma(4m)=1 => sigma(2phi( m))/7sigma( m)=1 => sigma(2phi( m))/ sigma( m)=7. So we get the condition sigma(phi( m)) / sigma( m)=9 <=> sigma(2phi( m))/ sigma( m)=7 which will fail. So if there is a (very) big odd number n in A066831 (numbers n such that sigma(n) divides sigma(phi(n))) with A066831(n) = 9, the conjecture is wrong. I admit I could not yet find such a number, nor do i really know it exists, i.e., A067385(9) exists. (End)

References

• R. K. Guy, Unsolved Problems in Number Theory, B42.

Links

• Harry J. Smith, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A033631, A067382, A067383, A067384, A067385.

Programs

Mathematica

For[ n=1, True, n++, If[ Mod[ DivisorSigma[ 1, EulerPhi[ n ] ], DivisorSigma[ 1, n ] ]==0, Print[ n ] ] ]
Select[Range[15000],Divisible[DivisorSigma[1, EulerPhi[#]], DivisorSigma[1,#]]&] (* Harvey P. Dale, Oct 19 2011 *)

PARI

isok(k) = { sigma(eulerphi(k)) % sigma(k) == 0 } \\ Harry J. Smith, Mar 30 2010

Extensions

More terms from Vladeta Jovovic and Robert G. Wilson v, Jan 20 2002

Edited by Dean Hickerson, Jan 20 2002