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.

1, 2, 588, 864, 2430, 7776, 27000, 55296, 69984, 82134, 215622, 432000, 497664, 629856, 675000, 862488, 1499136, 1749600, 2187000, 2667168, 3449952, 3538944, 4287500, 4312440, 4478976, 4563000, 5668704, 6912000, 10800000, 13045131

Offset: 1

Author

Tanya Khovanova, Aug 30 2006

Keywords

• nonn

Comments

Numbers k such that A000010(k) + A000203(k) is divisible by A007947(k)^2.

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

Examples

			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.
		

References

• J.-M. De Koninck, Those Fascinating Numbers, Amer. Math. Soc., 2009, page 71, entry 588.
• J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Ellipses, 2004, Problème 749, pp. 95 and 319.

Links

• Amiram Eldar, Table of n, a(n) for n = 1..145

Crossrefs

Cf. A000010, A000203, A007947.

This sequence is similar to A097982.

Programs

Mathematica

Do[If[IntegerQ[(DivisorSigma[1, n] + EulerPhi[n])/(Times @@ Transpose[FactorInteger[n]][[1]])^2], Print[n]], {n, 1, 1000000}]

PARI

isok(k) = (((eulerphi(k) + sigma(k)) % factorback(factorint(k)[, 1])^2) == 0); \\ Michel Marcus, Dec 03 2020

Extensions

a(17)-a(30) from Donovan Johnson, Feb 05 2010

a(1) = 1 inserted by Amiram Eldar, Aug 24 2019