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%I #8 May 16 2013 23:37:04
%S 7,13,19,37,73,97,109,163,193,369,433,487,577,769,1153,1297,1459,2593,
%T 2917,3457,3889,10369,12289,17497,18433,39367,52489,139969,147457,
%U 209953,331777,472393,629857,746497,786433,839809,995329,1179649,1492993,1769473
%N Numbers n such that sigma(n) = 2n - 3*phi(phi(n)).
%C 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.
%H Donovan Johnson, <a href="/A110074/b110074.txt">Table of n, a(n) for n = 1..70</a> (terms < 2*10^10)
%e 369 is in the sequence because 2*369-3*phi(phi(369))=546=13*42 =sigma(9)*sigma(41)=sigma(9*41)=sigma(369).
%t Do[If[DivisorSigma[1, m] == 2m - 3EulerPhi[EulerPhi[m]], Print[m]], {m, 1500000}]
%o (PARI) is(n)=sigma(n)==2*n-3*eulerphi(eulerphi(n)) \\ _Charles R Greathouse IV_, May 15 2013
%Y Cf. A110073.
%K nonn
%O 1,1
%A _Farideh Firoozbakht_, Jul 25 2005