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a(n) is the smallest number m greater than 1 such that phi(m) = d(m)^n, where d(m) is number of positive divisors of m; if there is no such m, a(n)=1.
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%I #21 Jun 15 2020 21:16:49

%S 3,5,85,17,1285,4369,559876,257,327685,1114129,1114521441417,16843009,

%T 160490068541289,1925878801139721,23110536763219977,65537,

%U 3327917287071744009,39934999967815157769,479219999336720898057,5750639996603165650953,69007679885506346588169,828092158571811231498249,9937105900443065378930697

%N a(n) is the smallest number m greater than 1 such that phi(m) = d(m)^n, where d(m) is number of positive divisors of m; if there is no such m, a(n)=1.

%C For n=0,1,2,3, and 4, a(2^n) = A000215(n), the n-th Fermat prime.

%C Conjecture: A000005(a(n)) <= 12 for all n. [_Max Alekseyev_, May 07 2010]

%C This conjecture holds throughout the first 102 terms. - _David A. Corneth_, Jun 14 2020

%H David A. Corneth, <a href="/A107655/b107655.txt">Table of n, a(n) for n = 1..100</a>

%H Max Alekseyev, <a href="https://home.gwu.edu/~maxal/gpscripts/invphi.gp">PARI scripts for various problems</a> (see invphitau there).

%e a(10) = 1114129 because phi(1114129) = d(1114129)^10 and 1114129 is the smallest number m greater than 1 that phi(m) = 1048576 = 4^10 = d(m)^10.

%o (PARI) a(n)=res = oo; for(i=2, oo, if(i^n > res, return(res)); c=invphitau(i^n,i); if(#c>0, res=c[1])) \\ for invphitau, see Alekseyev link \\ _David A. Corneth_, Jun 14 2020

%Y Cf. A000005, A000215, A033844.

%K nonn

%O 1,1

%A _Farideh Firoozbakht_, Jun 06 2005

%E Terms a(11) onward from _Max Alekseyev_, May 07 2010

%E Terms a(20)-a(23), offset corrected by _David A. Corneth_, Jun 14 2020