OFFSET
1,2
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
LINKS
Max Alekseyev, Table of n, a(n) for n = 1..614 (first 470 terms from Enrique Pérez Herrero)
EXAMPLE
a(2) = A107655(3) = 85.
MATHEMATICA
Select[Range[132000], EulerPhi[#]==DivisorSigma[0, #]^3&] (* Harvey P. Dale, Dec 28 2022 *)
PROG
(PARI) isok(m) = eulerphi(m) == numdiv(m)^3; \\ Michel Marcus, Oct 18 2019
CROSSREFS
KEYWORD
nonn,fini,full
AUTHOR
Benoit Cloitre, Mar 25 2002
STATUS
approved