%I #26 Aug 18 2024 20:15:40
%S 1,13,71,87,89,181,203,305,319,362,667,899,1257,1363,1421,1525,1711,
%T 1798,1889,2407,2501,2933,3103,4609,4615,4687,4843,5002,5191,6583,
%U 7123,7625,7627,9374,9947,10063,10411,10991,11107,12989,13543,13891,14587
%N Numbers n such that sigma(n) divides sigma(phi(n)).
%C For odd n, if sigma(phi(n))/sigma(n)=3 then sigma(phi(2*n))/sigma(2*n)=1. - _Vladeta Jovovic_, Jan 21 2002.
%C Comments from _Vim Wenders_, Nov 01 2006: (Start)
%C 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.
%C 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)
%D R. K. Guy, Unsolved Problems in Number Theory, B42.
%H Harry J. Smith, <a href="/A066831/b066831.txt">Table of n, a(n) for n = 1..1000</a>
%t For[ n=1, True, n++, If[ Mod[ DivisorSigma[ 1, EulerPhi[ n ] ], DivisorSigma[ 1, n ] ]==0, Print[ n ] ] ]
%t Select[Range[15000],Divisible[DivisorSigma[1, EulerPhi[#]], DivisorSigma[1,#]]&] (* _Harvey P. Dale_, Oct 19 2011 *)
%o (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
%Y Cf. A033631, A067382, A067383, A067384, A067385.
%K nonn
%O 1,2
%A _Benoit Cloitre_, Jan 19 2002
%E More terms from _Vladeta Jovovic_ and _Robert G. Wilson v_, Jan 20 2002
%E Edited by _Dean Hickerson_, Jan 20 2002