%I
%S 275,295,2995,299995,2999995,278736495,299999995,299999999995
%N Numbers n such that reverse(n) = phi(n) + sigma(n).
%C For n>0 5*(6*10^A056716(n)1) is in this sequence. In fact if p = 6*10^n1 is prime and n>0 (p>5) then m = 5*p is in the sequence. That's because phi(m) = phi(5*p) = 4*(6*10^n2) = 24*10^n8 and sigma(m)= 6*6*10^n, so phi(m) + sigma(m) = 6*10^(n+1)8 = 5.(9)(n).2 = reversal(2.(9)(n).5) = reversal (3*10^(n+1)5) = reversal(m)(dot between numbers means concatenation and "(9)(n)" means number of 9's is n). For example 299999995 is in the sequence because 6*10^71 is prime and 299999995 = 5*(6*10^71); 299999999995 is in sequence because 6*10^101 is prime and 299999999995 = 5*(6*10^101). Next term is greater than 80000000.  _Farideh Firoozbakht_, Jan 11 2005
%C Next term is greater than 10^9.  _Farideh Firoozbakht_, Jan 23 2005
%C a(9) > 10^13.  _Giovanni Resta_, Feb 08 2014
%e Reverse(275) = 572 = 200 + 372 = phi(275) + sigma(275).
%t Select[Range[10^6], FromDigits[Reverse[IntegerDigits[ # ]]] == EulerPhi[ # ] + DivisorSigma[1, # ] &]
%Y Cf. A056716, A102284.
%K base,nonn
%O 1,1
%A _Joseph L. Pe_, May 12 2002
%E One more term from _Farideh Firoozbakht_, Jan 11 2005
%E More terms from _Farideh Firoozbakht_, Jan 23 2005
%E a(8) from _Giovanni Resta_, Nov 03 2012
