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A230032
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Numbers n such that phi(sigma(n)) + sigma(phi(n)) < n.
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2
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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
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1,1
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COMMENTS
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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 (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)
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LINKS
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MATHEMATICA
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Do[If[EulerPhi[DivisorSigma[1, n]] + DivisorSigma[1, EulerPhi[n]] < n, Print[n]], {n, 3300000}]
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PROG
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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