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Numbers n such that sigma(n)+phi(n) has exactly 4 divisors.
3

%I #14 Aug 13 2019 11:23:22

%S 3,5,6,7,10,11,13,16,17,19,22,23,25,27,29,31,37,40,41,43,46,47,52,53,

%T 58,59,61,64,67,68,71,72,73,79,80,82,83,89,97,98,101,103,106,107,109,

%U 113,117,127,128,131,136,137,139,144,149,151,157,162,163,166,167,169

%N Numbers n such that sigma(n)+phi(n) has exactly 4 divisors.

%C For all terms up to 10^12, sigma(n)+phi(n) is a product of 2 distinct primes. The only other possibility is that sigma(n)+phi(n) is a cube of a prime, for some n which is either a square or twice a square; does this occur? If not, then this sequence is contained in A067351.

%H Amiram Eldar, <a href="/A067350/b067350.txt">Table of n, a(n) for n = 1..10000</a>

%F A000005(A000010(n) + A000203(n)) = A067349(n) = 4.

%e Includes all odd primes and some composites; e.g. 22 and 25, since sigma(22)+phi(22)=36+10=46=2*23 and sigma(25)+phi(25)=31+20=51=3*17.

%t Select[ Range[ 1, 200 ], DivisorSigma[ 0, DivisorSigma[ 1, # ]+EulerPhi[ # ] ]==4& ]

%o (PARI) isok(n) = numdiv(sigma(n)+eulerphi(n)) == 4; \\ _Michel Marcus_, Aug 13 2019

%Y Cf. A000005, A000010, A000203, A067349, A067351.

%K nonn

%O 1,1

%A _Labos Elemer_, Jan 17 2002

%E Edited by _Dean Hickerson_, Jan 20 2002