A069282 - OEIS

%I #8 Feb 11 2014 19:05:28

%S 1,2,3,4,5,6,7,8,9,535,767,2110,6188,6991,8299,8816,9928,13580,20502,

%T 21100,48991,50805,53035,58085,58585,59395,69991,82428,82770,88188,

%U 135800,211000,827700,1358000,2110000,2753010,3269623,5808085,5846485

%N Numbers n such that phi(reversal(n)) = reversal(phi(n)). Ignore leading 0's.

%C For an arithmetical function f, call the arguments n such that f(reverse(n)) = reverse(f(n)) the "palinpoints" of f. This sequence is the sequence of palinpoints of f(n) = phi(n).

%C If n is in the sequence and 10 divides n then for each natural number m, 10^m*n is in the sequence because phi(reversal(10^m*n))=phi(reversal(n))=reversal(phi(n)) =reversal(10^m*phi(n))=reversal(phi(10^m*n)). This sequence is infinite because the numbers 2110,13580,82770,8415570 are in the sequence and according to the above assertion all numbers of the form 2110*10^m, 13580*10^m, 82770*10^m & 8415570*10^m are in the sequence. If n is in the sequence and 10 doesn't divide n then it is obvious that the reversal of n is also in the sequence. If both numbers 5*10^(n-1)-1 & 7*10^n-9 are prime we can easily show that 7*10^n-9 is in the sequence. - _Farideh Firoozbakht_, Jan 18 2006

%C If both numbers 49*10^n-9 & 125*10^(n-2)-1 are prime then 49*10^n-9 is in the sequence (the proof is easy). 3 is the only known number n such that both numbers 49*10^n-9 & 125*10^(n-2)-1 are primes. - _Farideh Firoozbakht_, Jan 26 2006

%e Let f(n) = phi(n). Then f(6188) = 2304, f(8816) = 4032, so f(reverse(6188)) = reverse(f(6188)). Therefore 6188 belongs to the sequence.

%t rev[n_] := FromDigits[Reverse[IntegerDigits[n]]]; f[n_] := EulerPhi[n]; Select[Range[10^6], f[rev[ # ]] == rev[f[ # ]] &]

%K base,nonn

%O 1,2

%A _Joseph L. Pe_, Apr 15 2002

%E More terms from _Farideh Firoozbakht_, Jan 18 2006