%I
%S 1,21,63,270,291,2991,6102,46676013,69460293,2346534651,6313047393,
%T 23400000651,80050617822,234065340651,234659934651,2340000000651,
%U 2530227348360,2934000006591
%N Numbers n such that phi(n) = reversal(n).
%C If 10^n3 is prime (n is in the sequence A089765) and m=3*(10^n3) then m is in this sequence, for example 299999999999999991 is a term of this sequence because 299999999999999991=3*(10^173) and 17 is in the sequence A089675. So 3*(10^A0896753) is a subsequence of this sequence, A101700 is this subsequence.  _Farideh Firoozbakht_, Dec 26 2004
%C A072395 is a subsequence of this sequence. If m is in the sequence and 10 doesn't divide m then reversal(m) is in the sequence A085331, so see Comments on A085331.  _Farideh Firoozbakht_, Jan 09 2005
%C If p=(79*10^(4n+1)83)/101 is prime then 3p is in the sequence. The proof is easy. 21, 2346534651 & 3*(79*10^269783)/101 are the first three such terms.  _Farideh Firoozbakht_, Apr 22 2008, Aug 16 2008
%C a(19) > 10^13.  _Giovanni Resta_, Aug 07 2019
%e phi(291) = 192.
%e phi(6102) = 2016 = reversal(6102), so 6102 belongs to the sequence.
%t Do[If[EulerPhi[n] == FromDigits[Reverse[IntegerDigits[n]]], Print[n]], {n, 1, 10^5}]
%o (PARI) for( n=1,1e9, A004086(n)==eulerphi(n) & print1(n","))
%Y Cf. A101700, A004086, A000010, A085331, A072395, A101700, A102278.
%K nonn,base,hard,more
%O 1,2
%A _Joseph L. Pe_, Apr 11 2002
%E More terms from _Farideh Firoozbakht_, Aug 31 2004
%E One more term from _Farideh Firoozbakht_, Jan 09 2005
%E a(11)a(13) from _Donovan Johnson_, Feb 03 2012
%E a(14)a(15) from _Giovanni Resta_, Oct 28 2012
%E a(16)a(18) from _Giovanni Resta_, Aug 07 2019
