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%I #6 Mar 11 2014 01:32:26
%S 0,1,1,0,1,0,0,1,1,0,0,1,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,1,0,1,0,1,1,1,
%T 0,0,0,1,0,0,1,0,1,0,1,1,0,1,1,0,0,1,0,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1,
%U 0,0,1,0,1,0,0,1,0,1,1,0,1,1,0,0,1,0,0,0,1,0,1,0,0,0,1,0,1,1,0,0,0,0,1,1,1
%N a(0)=0. For n >=1, a(n) = 0 if the binary representation of n occurs at least once in the concatenation of (a(0),a(1),...,a(n-1)). a(n) = 1 otherwise.
%e The binary representation of 20 is 10100. This occurs in the concatenation of terms a(0) through a(19) like so: 01(10100)1100100111100. So a(20) = 0.
%t f[l_List]:=Append[l,Boole[StringPosition[ToString[FromDigits[l]],ToString[FromDigits[IntegerDigits[Length[l],2]]]]=={}]];Nest[f,{0},125] [From _Ray Chandler_, Nov 09 2008]
%Y Cf. A143220, A143221.
%K base,nonn
%O 0,1
%A _Leroy Quet_, Jul 30 2008
%E Extended by _Ray Chandler_, Nov 09 2008
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