A068560 Numbers k such that phi(k) = tau(k)^2.

1, 5, 34, 63, 76, 128, 136, 170, 315, 364, 380, 444, 640, 680, 972, 1820, 1824, 1836, 2142, 2220, 4788, 4860, 6000, 8064, 8568, 8736, 9120, 9180, 10710, 23940, 40320, 42840, 43680

Offset: 1

Author

Benoit Cloitre, Mar 25 2002

Keywords

• fini
• full
• nonn

Comments

This sequence is finite because phi(k) >= sqrt(k) for all k >= 6, and for any e > 0, tau(k) < k^e for k large enough. Choosing e=1/4 gives tau(k)^2 < sqrt(k) <= phi(k). It remains unknown, however, if this sequence is full. - Nathaniel Johnston, Apr 28 2011

It can be shown that tau(k) <= 120 and the sequence is complete. - Max Alekseyev, May 30 2024

Examples

			a(2) = A107655(2) = 5.
		

Crossrefs

Cf. A039770, A068559, A107655, A114063.

Programs

Mathematica

Do[If[EulerPhi[n] == DivisorSigma[0, n]^2, Print[n]], {n, 10^5}] (* Ryan Propper, Jun 09 2006 *)
Select[Range[10^5], EulerPhi[#] == DivisorSigma[0, #]^2 &] (* Alonso del Arte, Aug 25 2011 *)

Extensions

More terms from Ryan Propper, Jun 09 2006

"full" keyword added by Max Alekseyev, May 30 2024