%I #17 Jan 12 2023 07:33:15
%S 1,2,3,4,6,8,10,12,14,16,18,20,24,28,32,36,40,48,54,56,64,72,80,96,
%T 108,112,128,144,160,162,192,216,224,256,288,320,324,384,432,448,486,
%U 512,576,640,648,768,864,896,972,1024,1152,1280,1296,1458,1536,1728,1792,1944,2048,2304,2560
%N Numbers k such that phi(phi(k)) divides k, where phi(k) is A000010(k).
%C 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.
%H Giovanni Resta, <a href="/A283808/b283808.txt">Table of n, a(n) for n = 1..10000</a>
%H M. Hausman, <a href="http://dx.doi.org/10.4153/CMB-1982-015-x">The solution of a special arithmetic equation</a>, Canad. Math. Bull, 1982, 25(1), 114-117.
%e 56 is in the sequence because phi(phi(56)) = 8 divides 56.
%t Select[Range[1000], Mod[#, EulerPhi@ EulerPhi@ #] == 0 &]
%o (PARI) alias(e, eulerphi);
%o for(n = 1, 1000, if(!Mod(n,e(e(n))), print1(n,", "))) \\ _Indranil Ghosh_, Mar 18 2017
%o (Python)
%o from sympy import totient as e
%o print([n for n in range(1, 1001) if n%e(e(n))==0]) # _Indranil Ghosh_, Mar 18 2017
%Y Cf. A010554, A007694, A000010, A019278.
%K nonn
%O 1,2
%A _Giovanni Resta_, Mar 17 2017