Search: id:A104145
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%I A104145
%S A104145 1,0,2,1,2,1,3,2,0,1,1,0,1,0,2,1,2,1,3,2,3,2,4,3,1,0,2,1,2,1,3,2,0,1,1,
               0,1,0,2,1,1,
%T A104145 2,0,1,0,1,1,0,1,0,2,1,2,1,3,2,0,1,1,0,1,0,2,1,0,1,1,0,1,0,2,1,1,2,0,1,
               0,1,1,
%U A104145 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
%V A104145 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,-1,
               1,0,1,0,2,1,-1,
%W A104145 -2,0,-1,0,-1,1,0,1,0,2,1,2,1,3,2,0,-1,1,0,1,0,2,1,0,-1,1,0,1,0,2,1,-1,
               -2,0,-1,0,-1,1,
%X A104145 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 A104145 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.
%H A104145 Leroy Quet, <a href="http://www.prism-of-spirals.net/">Home Page</a> 
               (listed in lieu of email address)
%F A104145 a(n) = 1 - A137412(n). - Leroy Quet, Apr 22 2008
%e A104145 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 A104145 Cf. A137412.
%Y A104145 Sequence in context: A137753 A134780 A154819 this_sequence A123675 A123400 
               A023116
%Y A104145 Adjacent sequences: A104142 A104143 A104144 this_sequence A104146 A104147 
               A104148
%K A104145 easy,sign
%O A104145 1,3
%A A104145 Leroy Quet, Mar 07 2005
%E A104145 More terms from Joshua Zucker (joshua.zucker(AT)stanfordalumni.org), 
               May 10 2006

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