A283808 Numbers k such that phi(phi(k)) divides k, where phi(k) is A000010(k).

1, 2, 3, 4, 6, 8, 10, 12, 14, 16, 18, 20, 24, 28, 32, 36, 40, 48, 54, 56, 64, 72, 80, 96, 108, 112, 128, 144, 160, 162, 192, 216, 224, 256, 288, 320, 324, 384, 432, 448, 486, 512, 576, 640, 648, 768, 864, 896, 972, 1024, 1152, 1280, 1296, 1458, 1536, 1728, 1792, 1944, 2048, 2304, 2560

Offset: 1

Author

Giovanni Resta, Mar 17 2017

Keywords

• nonn

Comments

M. Hausman has proved (see Links) that a number belongs to this sequence if and only if it is of one of the following forms: 2^s, 2^s * 3^t, 5 * 2^t, or 7 * 2^t , where s >= 0 and t >= 1.

Examples

			56 is in the sequence because phi(phi(56)) = 8 divides 56.
		

Links

• Giovanni Resta, Table of n, a(n) for n = 1..10000
• Miriam Hausman, The solution of a special arithmetic equation, Canad. Math. Bull., 25(1) (1982), 114-117.

Crossrefs

Cf. A010554, A007694, A000010, A019278.

Programs

Mathematica

Select[Range[1000], Mod[#, EulerPhi@ EulerPhi@ #] == 0 &]

PARI

alias(e, eulerphi);
for(n = 1, 1000, if(!Mod(n,e(e(n))), print1(n,", "))) \\ Indranil Ghosh, Mar 18 2017

Python

from sympy import totient as e
print([n for n in range(1, 1001) if n%e(e(n))==0]) # Indranil Ghosh, Mar 18 2017

Formula

Sum_{n>=1} 1/a(n) = 667/210. - Amiram Eldar, Dec 13 2024