%I
%S 37,397,1853,15503,48776,198683,200882,1061361,3542805,3564217,
%T 3868867,3962197,4438616,19844683,198444683,202195682,309520655,
%U 431092646,439419646,500729929,535973599,3566790217,3963436297,4149753226,17296101143,39560402197
%N Numbers n such that phi(n)=reversal(n)n.
%C All primes of the form 4*10^n3 are in the sequence because if 4*10^n3 is prime then phi(4*10^n3)=(4*10^n4) =(8*10^n7)(4*10^n3)=reversal(4*10^n3)(4*10^n3).
%C Also if n>1 and p=(94*10^n+113)/9 is prime then 19*p is in the sequence (the proof is easy). Next term is greater than 125*10^6.
%C If p=(1/303)*(232*10^(4n)+71) is prime then 7*p is in the sequence (the proof is easy). The first four such terms happen for n=2, 101, 104 & 444 and numbers of digits of these terms of the sequence are 9, 405, 417 & 1777 respectively.  _Farideh Firoozbakht_, Jan 02 2008
%C a(32) > 10^12.  _Giovanni Resta_, Oct 28 2012
%H Giovanni Resta, <a href="/A115926/b115926.txt">Table of n, a(n) for n = 1..31</a>
%e If n=37, phi(37)= 36 = 7337
%t Do[If[EulerPhi[n]==FromDigits[Reverse[IntegerDigits[n]]]n, Print[n]],{n,125000000}]
%t Do[If[EulerPhi[n]==FromDigits[Reverse[IntegerDigits[n]]]n, Print[n]], {n, 600000000}]  Jessica M. Cornwall (jmc510(AT)psu.edu), Apr 05 2006
%Y Cf. A072393.
%K more,nonn,base
%O 1,1
%A _Farideh Firoozbakht_, Jan 31 2006
%E More terms from Jessica M. Cornwall (jmc510(AT)psu.edu), Apr 05 2006
%E a(22)a(31) from _Giovanni Resta_, Oct 28 2012
