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A run of 3*2^n 0's followed by a run of 3*2^n 1's, for n=0, 1, 2, ...
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%I #7 Jun 01 2012 16:05:52

%S 0,0,0,1,1,1,0,0,0,0,0,0,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,

%T 1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,

%U 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0

%N A run of 3*2^n 0's followed by a run of 3*2^n 1's, for n=0, 1, 2, ...

%F a(n) = (1 - (-1)^A079944(A002264(n)) )/2, A079944(A002264(n))=floor(log[2](4*(floor((n+6)/3))/3)) - floor(log[2](floor((n+6)/3))) - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 24 2003

%F Also a(n) = A079944(A002264(n)) = floor(log[2](4*(floor((n+6)/3))/3)) - floor(log[2](floor((n+6)/3))) - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 24 2003

%p f := (c,n)->seq(c,i = 1..3*2^n); [f(0,0),f(1,0),f(0,1),f(1,1),f(0,2),f(1,2),f(0,3),f(1,3)]; f;

%t Flatten[Table[{PadRight[{},3*2^n,0],PadRight[{},3*2^n,1]},{n,0,4}]] (* _Harvey P. Dale_, Jun 01 2012 *)

%Y Equals A080586 - 1.

%K nonn

%O 0,1

%A _N. J. A. Sloane_, Feb 23 2003