OFFSET
1,1
COMMENTS
Let ps(n) be number of terms of the sequence up to n, it seems that ps(n) ~ n/100000. Is it true that 6 divides each term of the sequence?
I guess that there is no number n such that phi(sigma(n)) + sigma(phi(n)) = n.
From M. F. Hasler, Oct 31 2013: (Start)
Most terms of the sequence are of the form given in the following
Theorem: If p is a safe prime (A005385), then n = 6p is a term of this sequence if and only if (1-1/q1)*...*(1-1/qr) + 7/12 < p/(p+1), where q1,...,qr are the distinct odd prime factors of p+1.
Proof: Write p+1 = 2^a 3^b Q with gcd(Q,6)=1 and assume (p-1)/2 is prime. For n = 6p, an easy calculation yields phi(sigma(n)) + sigma(phi(n)) = n*(1+1/p)*(2/3*(1-1/q2)*...*(1-1/qr)+7/12), where q2,...,qr are the prime factors of Q. #
Corollary: n=6p is in the sequence when p is a safe prime and p+1 is a multiple of 2*3*5*7*11 or of 2*3*5*7*13*q with some prime q>13, q<80. (End)
LINKS
Donovan Johnson, Table of n, a(n) for n = 1..1000
MATHEMATICA
Do[If[EulerPhi[DivisorSigma[1, n]] + DivisorSigma[1, EulerPhi[n]] < n, Print[n]], {n, 3300000}]
PROG
(PARI) is_A230032(n)={eulerphi(sigma(n))+sigma(eulerphi(n))<n} \\ M. F. Hasler, Oct 31 2013
CROSSREFS
KEYWORD
nonn
AUTHOR
Farideh Firoozbakht, Oct 28 2013
STATUS
approved