OFFSET
1,1
COMMENTS
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).
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.
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
a(32) > 10^12. - Giovanni Resta, Oct 28 2012
LINKS
Giovanni Resta, Table of n, a(n) for n = 1..31
EXAMPLE
If n=37, phi(37)= 36 = 73-37
MATHEMATICA
Do[If[EulerPhi[n]==FromDigits[Reverse[IntegerDigits[n]]]-n, Print[n]], {n, 125000000}]
Do[If[EulerPhi[n]==FromDigits[Reverse[IntegerDigits[n]]]-n, Print[n]], {n, 600000000}] - Jessica M. Cornwall (jmc510(AT)psu.edu), Apr 05 2006
CROSSREFS
KEYWORD
more,nonn,base
AUTHOR
Farideh Firoozbakht, Jan 31 2006
EXTENSIONS
More terms from Jessica M. Cornwall (jmc510(AT)psu.edu), Apr 05 2006
a(22)-a(31) from Giovanni Resta, Oct 28 2012
STATUS
approved