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

%I #6 May 04 2014 10:42:41

%S 1,1,1,2,2,2,1,1,1,1,1,1,2,2,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,

%T 2,2,2,2,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,

%U 2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1

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

%F a(n) = ( 3 - (-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) = 1+A079944(A002264(n))=floor(log[2](8*(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(1,0),f(2,0),f(1,1),f(2,1),f(1,2),f(2,2),f(1,3),f(2,3)]; f;

%t Flatten[Table[{PadRight[{},3*2^n,1],PadRight[{},3*2^n,2]},{n,0,4}]] (* _Harvey P. Dale_, May 04 2014 *)

%Y Cf. A079948, A080587.

%Y Equals 1 + A080584.

%K nonn

%O 0,4

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