OFFSET
1,1
COMMENTS
If m>1 and p=2*10^m+3 is prime then n=27*p is in the sequence because n-reversal(n)=27*(2*10^m+3)-reversal(27*(2*10^m+3))= (54*10^m+81)-(18*10^m+45)=36*10^m+36=18*(2*10^m+2)=phi(27)* phi(2*10^m+3)=phi(27*(2*10^m+3))=phi(n). Also if m>2 and p=(389*10^m+109)/3 is prime then 7*p is in the sequence (the proof is easy). Next term is greater than 2*10^8. - Farideh Firoozbakht, Jan 27 2006
a(51) > 10^12. - Giovanni Resta, Oct 28 2012
LINKS
Giovanni Resta, Table of n, a(n) for n = 1..50
EXAMPLE
91 - 19 = 72 = phi(91), so 91 is a term of the sequence.
MATHEMATICA
Select[Range[10^5], # - FromDigits[Reverse[IntegerDigits[n]]] == EulerPhi[ # ] &]
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Joseph L. Pe, Jul 21 2002
EXTENSIONS
More terms from Farideh Firoozbakht, Jan 27 2006
a(22)-a(29) from Donovan Johnson, Dec 04 2011
STATUS
approved