OFFSET
1,1
COMMENTS
If p>3 and 2^p-1 is prime (a Mersenne prime) then 35*2^(p-2)*(2^p-1) is in the sequence. So 35*2^(A000043-2)*(2^A000043-1) is a subsequence of this sequence.
If p>2 and 2^p-1 is prime (a Mersenne prime) then 3*2^(p-2)*(2^p-1) is in the sequence (the proof is easy). - Farideh Firoozbakht, Dec 23 2007
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000 (calculated using data from Jud McCranie, terms 1..1000 from Donovan Johnson)
Kevin A. Broughan and Daniel Delbourgo, On the Ratio of the Sum of Divisors and Euler’s Totient Function I, Journal of Integer Sequences, Vol. 16 (2013), Article 13.8.8.
Kevin A. Broughan and Qizhi Zhou, On the Ratio of the Sum of Divisors and Euler's Totient Function II, Journal of Integer Sequences, Vol. 17 (2014), Article 14.9.2.
EXAMPLE
p>3, q=2^p-1(q is prime); m=35*2^(p-2)*q so sigma(m)=48*(2^(p-1)-1)*2^p=8*(24*2^(p-3)*(2^p-2))=8*phi(m) hence m is in the sequence.
sigma(553000) = 1497600 = 8*187200 = 8*phi(553000) so 553000 is in the sequence but 553000 is not of the form 35*2^(p-2)*(2^p-1).
MATHEMATICA
Do[If[DivisorSigma[1, m] == 8*EulerPhi[m], Print[m]], {m, 1000000}]
Select[Range[800000], DivisorSigma[1, #]==8*EulerPhi[#]&] (* Harvey P. Dale, Sep 12 2018 *)
PROG
(PARI) is(n)=sigma(n)==8*eulerphi(n) \\ Charles R Greathouse IV, May 09 2013
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Farideh Firoozbakht, Apr 01 2005
STATUS
approved