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%I #9 Mar 11 2014 01:32:15
%S 1,0,2,1,2,1,3,2,0,-1,1,0,1,0,2,1,2,1,3,2,3,2,4,3,1,0,2,1,2,1,3,2,0,
%T -1,1,0,1,0,2,1,-1,-2,0,-1,0,-1,1,0,1,0,2,1,2,1,3,2,0,-1,1,0,1,0,2,1,
%U 0,-1,1,0,1,0,2,1,-1,-2,0,-1,0,-1,1,0,1,0,2,1,2,1,3,2,0,-1,1,0,1,0,2,1,-1,-2,0,-1,0,-1,1,0,-2,-3,-1,-2,-1,-2,0,-1,0,-1,1
%N a(1) = 1; let A(k) = sequence of first 2^(k-1) terms; then A(k+1) is concatenation of A(k) and (A(k)-1) if a(k) is odd, or concatenation of A(k) and (A(k)+1) if a(k) is even.
%F a(n) = 1 - A137412(n). - _Leroy Quet_, Apr 22 2008
%e a(3) = 2 is even, so A(4) (1,0,2,1,2,1,3,2), the first 8 terms of the sequence, is A(3) (1,0,2,1) concatenated with each term of A(3) plus one (2,1,3,2).
%Y Cf. A137412.
%K easy,sign
%O 1,3
%A _Leroy Quet_, Mar 07 2005
%E More terms from _Joshua Zucker_, May 10 2006
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