OFFSET
1,1
COMMENTS
Each prime number p of the form 2^k*3^j+1 where k & j are natural numbers is in the sequence because 2p-3*phi(phi(p))=2p-3*phi (2^k*3^j)=2p-3*(1-1/2)*(1-1/3)*2^k*3^j=2p-2^k*3^j=p+1=sigma(p). Conjecture: The sequence is infinite and 369 is the only composite term. I checked the validity of this conjecture up to 1.5*10^9.
LINKS
Donovan Johnson, Table of n, a(n) for n = 1..70 (terms < 2*10^10)
EXAMPLE
369 is in the sequence because 2*369-3*phi(phi(369))=546=13*42 =sigma(9)*sigma(41)=sigma(9*41)=sigma(369).
MATHEMATICA
Do[If[DivisorSigma[1, m] == 2m - 3EulerPhi[EulerPhi[m]], Print[m]], {m, 1500000}]
PROG
(PARI) is(n)=sigma(n)==2*n-3*eulerphi(eulerphi(n)) \\ Charles R Greathouse IV, May 15 2013
CROSSREFS
KEYWORD
nonn
AUTHOR
Farideh Firoozbakht, Jul 25 2005
STATUS
approved