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%I #9 Dec 17 2013 11:28:35
%S 20,30,200,300,2000,3000,20000,30000,200000,300000,2000000,3000000,
%T 20000000,30000000
%N Numbers n such that omega(n) = digit-reverse(n).
%C omega(n) = A001221(n) is the number of distinct primes dividing n.
%F {n: A001221(n) = A004086(n)}
%e 300 is in the sequence because omega(300) = reversal(300) = 3.
%t Do[If[Length[ FactorInteger[ n ]]==FromDigits[Reverse[IntegerDigits[n]]], Print[n]],
%t {n, 1, 10^9}]
%t Select[Range[10^9],PrimeNu[#]==FromDigits[Reverse[IntegerDigits[#]]]&] (* or *) Flatten[Table[{2*10^n, 3*10^n}, {n, 10}]] (* _Harvey P. Dale_, Dec 17 2013 *)
%Y Cf. A001221.
%K nonn,base
%O 1,1
%A _Michel Lagneau_, Nov 18 2010
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