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A121850 Numbers k such that (phi(k) + sigma(k))/rad(k)^2 is an integer, that is (phi(k) + sigma(k)) is divisible by every prime factor of k squared.

%I A121850 #37 Dec 03 2020 19:21:14
%S A121850 1,2,588,864,2430,7776,27000,55296,69984,82134,215622,432000,497664,
%T A121850 629856,675000,862488,1499136,1749600,2187000,2667168,3449952,3538944,
%U A121850 4287500,4312440,4478976,4563000,5668704,6912000,10800000,13045131
%N A121850 Numbers k such that (phi(k) + sigma(k))/rad(k)^2 is an integer, that is (phi(k) + sigma(k)) is divisible by every prime factor of k squared.
%C A121850 Numbers k such that A000010(k) + A000203(k) is divisible by A007947(k)^2.
%C A121850 This sequence is infinite because every integer m = 32 * 3^(2r+1), r>=1 is a term: 864, 7776, 69984, ... (see De Koninck & Mercier reference). - _Bernard Schott_, Dec 02 2020
%D A121850 J.-M. De Koninck, Those Fascinating Numbers, Amer. Math. Soc., 2009, page 71, entry 588.
%D A121850 J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Ellipses, 2004, Problème 749, pp. 95 and 319.
%H A121850 Amiram Eldar, <a href="/A121850/b121850.txt">Table of n, a(n) for n = 1..145</a>
%e A121850 For example, phi(588) = 168, sigma(588) = 1596, 588 = 2^2*3*7^2. The product of all prime divisors is 42, its square is 1764. Hence phi(588) + sigma(588), which is equal to 1764 is divisible by the square of each prime divisor of 588.
%t A121850 Do[If[IntegerQ[(DivisorSigma[1, n] + EulerPhi[n])/(Times @@ Transpose[FactorInteger[n]][[1]])^2], Print[n]], {n, 1, 1000000}]
%o A121850 (PARI) isok(k) = (((eulerphi(k) + sigma(k)) % factorback(factorint(k)[, 1])^2) == 0); \\ _Michel Marcus_, Dec 03 2020
%Y A121850 Cf. A000010, A000203, A007947.
%Y A121850 This sequence is similar to A097982.
%K A121850 nonn
%O A121850 1,2
%A A121850 _Tanya Khovanova_, Aug 30 2006
%E A121850 a(17)-a(30) from _Donovan Johnson_, Feb 05 2010
%E A121850 a(1) = 1 inserted by _Amiram Eldar_, Aug 24 2019