OFFSET
1,1
COMMENTS
This sequence is a subsequence of A011257 because sqrt(phi(n)*sigma(n)) = 3*phi(n).
If 2^p-1 and 2*3^k-1 are two primes greater than 5 then n = 2^(p-2)*(2^p-1)*3^(k-1)*(2*3^k-1) (the product of two relatively prime terms 2^(p-2)*(2^p-1) and 3^(k-1)*(2*3^k-1) of A011257) is in the sequence. The proof is easy.
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.
MATHEMATICA
Select[Range[700000], DivisorSigma[1, # ]==9EulerPhi[ # ]&]
PROG
(PARI) is(n)=sigma(n)==9*eulerphi(n) \\ Charles R Greathouse IV, May 09 2013
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
M. F. Hasler and Farideh Firoozbakht, Aug 09 2009
STATUS
approved