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A115926 Numbers n such that phi(n)=reversal(n)-n. 1

%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^n-3 are in the sequence because if 4*10^n-3 is prime then phi(4*10^n-3)=(4*10^n-4) =(8*10^n-7)-(4*10^n-3)=reversal(4*10^n-3)-(4*10^n-3).

%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 = 73-37

%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

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Last modified January 26 23:05 EST 2020. Contains 331289 sequences. (Running on oeis4.)