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A110074
Numbers n such that sigma(n) = 2n - 3*phi(phi(n)).
2
7, 13, 19, 37, 73, 97, 109, 163, 193, 369, 433, 487, 577, 769, 1153, 1297, 1459, 2593, 2917, 3457, 3889, 10369, 12289, 17497, 18433, 39367, 52489, 139969, 147457, 209953, 331777, 472393, 629857, 746497, 786433, 839809, 995329, 1179649, 1492993, 1769473
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
Cf. A110073.
Sequence in context: A040034 A176229 A266268 * A058383 A005471 A249381
KEYWORD
nonn
AUTHOR
Farideh Firoozbakht, Jul 25 2005
STATUS
approved