%I #11 Oct 26 2020 15:16:00
%S 1,2,1,1,2,2,1,1,1,1,2,2,2,2,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,1,1,1,1,
%T 1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,1,1,1,1,1,1,
%U 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2
%N A run of 2^n 1's followed by a run of 2^n 2's, for n=0, 1, 2, ...
%C In the sequence of nonnegative integers (cf. A001477) substitute all n by 2^floor(n/2) occurrences of (1 + n mod 2); a(n)=A173920(n+2,3) for n>0. [From _Reinhard Zumkeller_, Mar 04 2010]
%H Robert Israel, <a href="/A079882/b079882.txt">Table of n, a(n) for n = 0..10000</a>
%F a(n) = floor(log[2](8*(n+2)/3)) - floor(log[2](n+2)). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 22 2003
%p f1 := n->[seq(1,i=1..2^n)]; f2 := n->[seq(2,i=1..2^n)]; s := []; for i from 0 to 10 do s := [op(s), op(f1(i)), op(f2(i))]; od: s;
%t Table[{PadRight[{},2^n,1],PadRight[{},2^n,2]},{n,0,5}]//Flatten (* _Harvey P. Dale_, Jul 22 2016 *)
%Y Partial sums give A079945. Equals 1 + A079944. Cf. A080584.
%Y First differences of A080637.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_, Feb 21 2003