

A230032


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


2



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
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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.
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 (11/q1)*...*(11/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 (p1)/2 is prime. For n = 6p, an easy calculation yields phi(sigma(n)) + sigma(phi(n)) = n*(1+1/p)*(2/3*(11/q2)*...*(11/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



MATHEMATICA

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


PROG



CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



