OFFSET
1,2
COMMENTS
For all large enough k, we have tau(k) < k^(1/5) and phi(k) > k^(4/5). Hence, tau(k)^4 < k^(4/5) < phi(k), implying that this sequence is finite. - Max Alekseyev, Mar 10 2016
Sequence is composed of 94030 terms. - Max Alekseyev, Jun 01 2024
LINKS
Max Alekseyev, Table of n, a(n) for n = 1..94030 (first 660 terms from Enrique Pérez Herrero; terms 661..7000 from Giovanni Resta)
EXAMPLE
phi(33003) = 20736. tau(33003) = 12, 20736 = 12^4.
a(2) = A107655(4) = 17.
MATHEMATICA
Select[Range[10^6], EulerPhi[#] == DivisorSigma[0, #]^4 &] (* Paolo Xausa, May 31 2024 *)
PROG
(PARI) isok(n) = eulerphi(n) == numdiv(n)^4; \\ Michel Marcus, Jan 22 2014
CROSSREFS
KEYWORD
fini,full,nonn
AUTHOR
Giovanni Resta, Feb 13 2006
STATUS
approved