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%I #51 Sep 08 2022 08:46:11
%S 4,6,12,15,20,21,28,30,33,39,42,44,51,52,57,66,68,69,76,78,87,92,93,
%T 102,111,114,116,123,124,129,138,141,148,159,164,172,174,177,183,186,
%U 188,201,212,213,219,222,236,237,244,246,249,258,267,268,282,284,291,292,303,309,316,318,321,327,332,339,354,356,366
%N Numbers n whose Euler's totient function phi(n), divided by two, plus one, p = (phi(n) / 2) + 1, is a divisor of n.
%C p is always a prime factor of n as well.
%C Except for the case n=6, p is always the greatest prime factor of n.
%C (n/p) is an upper bound on the rest of the prime factors 'q' of n, so always q <= (n/p).
%H Amiram Eldar, <a href="/A256241/b256241.txt">Table of n, a(n) for n = 1..10000</a>
%H David Morales Marciel, <a href="http://hobbymaths.blogspot.jp/2015/04/eulers-totient-function-statement.html">Euler's Totient function statement proposal</a>
%e For n = 4, phi(n) = 2, p = (phi(n)/2)+1 = 2, is prime and is a prime factor of 4.
%e For n = 6, phi(n) = 2, p = (phi(n)/2)+1 = 2, is prime and is a prime factor of 6.
%t aQ[n_] := Divisible[n, 1 + EulerPhi[n] / 2]; Select[Range[400], aQ] (* _Amiram Eldar_, Nov 06 2019 *)
%o (Python)
%o from sympy import totient
%o [n for n in range(1, 10**5) if n%((totient(n)/2)+1)==0]
%o (PARI) isok(n) = (n % (eulerphi(n)/2+1)) == 0; \\ _Michel Marcus_, Apr 20 2015
%o (Magma) [k:k in [1..370]| IsIntegral(k/(EulerPhi(k)/2+1))]; // _Marius A. Burtea_, Nov 06 2019
%Y Cf. A000010.
%K nonn
%O 1,1
%A _David Morales Marciel_, Apr 19 2015
%E Removed long Python code, and added very simple Python program (two lines) with sympy as suggested by the Editor.
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