1,2

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

Comments from Vim Wenders (vim(AT)gmx.li), 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)

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

For[ n=1, True, n++, If[ Mod[ DivisorSigma[ 1, EulerPhi[ n ] ], DivisorSigma[ 1, n ] ]==0, Print[ n ] ] ]

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

Sequence in context: A081860 A050403 A031442 this_sequence A067382 A158941 A128003

Adjacent sequences: A066828 A066829 A066830 this_sequence A066832 A066833 A066834

nonn

Benoit Cloitre (benoit7848c(AT)orange.fr), Jan 19 2002

More terms from Vladeta Jovovic (vladeta(AT)eunet.rs) and Robert G. Wilson v (rgwv(AT)rgwv.com), Jan 20 2002

Edited by Dean Hickerson (dean.hickerson(AT)yahoo.com), Jan 20 2002