|
| |
|
|
A066831
|
|
Numbers n such that sigma(n) divides sigma(phi(n)).
|
|
7
|
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
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.
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
|
|
|
|
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) { n=0; for (m=1, 10^10, if (sigma(eulerphi(m)) % sigma(m) == 0, write("b066831.txt", n++, " ", m); if (n==1000, return)) ) } \\ Harry J. Smith, Mar 30 2010]
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
