A068559 Numbers m such that phi(m) = tau(m)^3.

1, 85, 333, 436, 1542, 1875, 2907, 3285, 3488, 3796, 5196, 10280, 17532, 17776, 20080, 21250, 28305, 30368, 30555, 32708, 34748, 35308, 36860, 37060, 41544, 41568, 43068, 44004, 45162, 48468, 51930, 81324, 98304, 98688, 104856, 131070

Offset: 1

Author

Benoit Cloitre, Mar 25 2002

Keywords

• nonn
• fini
• full

Comments

For all large enough k, we have tau(k) < k^(1/4) and phi(k) > k^(3/4). Hence, tau(k)^3 < k^(3/4) < phi(k), implying that this sequence is finite. In fact, the sequence consists of 614 terms. - Max Alekseyev, May 30 2024

Examples

			a(2) = A107655(3) = 85.
		

Links

• Max Alekseyev, Table of n, a(n) for n = 1..614 (first 470 terms from Enrique Pérez Herrero)

Crossrefs

Subsequence of A039771. - Enrique Pérez Herrero, Aug 29 2010

Cf. A020488, A068560, A107655, A114063.

Programs

Mathematica

Select[Range[132000],EulerPhi[#]==DivisorSigma[0,#]^3&]  (* Harvey P. Dale, Dec 28 2022 *)

PARI

isok(m) = eulerphi(m) == numdiv(m)^3; \\ Michel Marcus, Oct 18 2019