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%I #9 Oct 02 2013 15:47:24
%S 111,1111,11111,1111111,11111111111,11111111111111111,
%T 11111111111111111111111111111111111111111111111,
%U 11111111111111111111111111111111111111111111111111111111111
%N Repunit semiprimes.
%F a(n) = A000042(A046413(n-1)). - _Ray Chandler_, Sep 06 2005
%e a(2)=1111 because 1111=11*101, so 1111 is semiprime as well as a repunit number.
%t Select[Table[FromDigits[PadRight[{},n,1]],{n,60}],PrimeOmega[#]==2&] (* _Harvey P. Dale_, Aug 28 2013 *)
%Y Cf. A046413 the repunit of length n has exactly 2 prime factors.
%K base,nonn
%O 1,1
%A _Shyam Sunder Gupta_, Feb 11 2005