cp's OEIS Frontend

A069215 Numbers n such that phi(n) = reversal(n).

%I A069215 #26 Aug 07 2019 04:17:54
%S A069215 1,21,63,270,291,2991,6102,46676013,69460293,2346534651,6313047393,
%T A069215 23400000651,80050617822,234065340651,234659934651,2340000000651,
%U A069215 2530227348360,2934000006591
%N A069215 Numbers n such that phi(n) = reversal(n).
%C A069215 If 10^n-3 is prime (n is in the sequence A089765) and m=3*(10^n-3) then m is in this sequence, for example 299999999999999991 is a term of this sequence because 299999999999999991=3*(10^17-3) and 17 is in the sequence A089675. So 3*(10^A089675-3) is a subsequence of this sequence, A101700 is this subsequence. - _Farideh Firoozbakht_, Dec 26 2004
%C A069215 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 A069215 If p=(79*10^(4n+1)-83)/101 is prime then 3p is in the sequence. The proof is easy. 21, 2346534651 & 3*(79*10^2697-83)/101 are the first three such terms. - _Farideh Firoozbakht_, Apr 22 2008, Aug 16 2008
%C A069215 a(19) > 10^13. - _Giovanni Resta_, Aug 07 2019
%e A069215 phi(291) = 192.
%e A069215 phi(6102) = 2016 = reversal(6102), so 6102 belongs to the sequence.
%t A069215 Do[If[EulerPhi[n] == FromDigits[Reverse[IntegerDigits[n]]], Print[n]], {n, 1, 10^5}]
%o A069215 (PARI) for( n=1,1e9, A004086(n)==eulerphi(n) & print1(n","))
%Y A069215 Cf. A101700, A004086, A000010, A085331, A072395, A101700, A102278.
%K A069215 nonn,base,hard,more
%O A069215 1,2
%A A069215 _Joseph L. Pe_, Apr 11 2002
%E A069215 More terms from _Farideh Firoozbakht_, Aug 31 2004
%E A069215 One more term from _Farideh Firoozbakht_, Jan 09 2005
%E A069215 a(11)-a(13) from _Donovan Johnson_, Feb 03 2012
%E A069215 a(14)-a(15) from _Giovanni Resta_, Oct 28 2012
%E A069215 a(16)-a(18) from _Giovanni Resta_, Aug 07 2019