A230032 Numbers n such that phi(sigma(n)) + sigma(phi(n)) < n.

138594, 249474, 277194, 471234, 554394, 665274, 900870, 1015554, 1081074, 1191954, 1244874, 1358274, 1385994, 1607754, 1801794, 1857234, 2189874, 2356170, 2356194, 2411634, 2439354, 2489754, 2522514, 2550234, 2633394, 2688834, 2702670, 2716554

Offset: 1

Author

Farideh Firoozbakht, Oct 28 2013

Keywords

• nonn

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

Crossrefs

Cf. A000010, A000203, A062401, A062402, A066950.

Programs

Mathematica

Do[If[EulerPhi[DivisorSigma[1,n]] + DivisorSigma[1,EulerPhi[n]] < n, Print[n]], {n,3300000}]

PARI

is_A230032(n)={eulerphi(sigma(n))+sigma(eulerphi(n))M. F. Hasler, Oct 31 2013