A066831 - OEIS

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

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OFFSET

1,2

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

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 *)

PROG

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

CROSSREFS

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

Sequence in context: A235454 A296831 A031442 * A067382 A340842 A253776