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Numbers k such that phi(k) = tau(k)^4, where tau(k) = A000005(k).
6

%I #35 Jun 02 2024 08:51:08

%S 1,17,514,8738,32301,33003,36351,41504,42292,43852,51860,62226,549117,

%T 561051,571311,599067,617967,629811,634005,657495,673184,674505,

%U 683168,701024,705568,718964,722684,732628,745484,759772,774368

%N Numbers k such that phi(k) = tau(k)^4, where tau(k) = A000005(k).

%C 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

%C Sequence is composed of 94030 terms. - _Max Alekseyev_, Jun 01 2024

%H Max Alekseyev, <a href="/A114063/b114063.txt">Table of n, a(n) for n = 1..94030</a> (first 660 terms from Enrique Pérez Herrero; terms 661..7000 from Giovanni Resta)

%e phi(33003) = 20736. tau(33003) = 12, 20736 = 12^4.

%e a(2) = A107655(4) = 17.

%t Select[Range[10^6], EulerPhi[#] == DivisorSigma[0, #]^4 &] (* _Paolo Xausa_, May 31 2024 *)

%o (PARI) isok(n) = eulerphi(n) == numdiv(n)^4; \\ _Michel Marcus_, Jan 22 2014

%Y Subsequence of A078164.

%Y Cf. A020488, A062516, A063469, A063470, A107655, A112954, A112955, A175667, A068560, A068559.

%K fini,full,nonn

%O 1,2

%A _Giovanni Resta_, Feb 13 2006