%I A037011
%S A037011 1,0,0,1,0,0,1,0,0,0,0,0,1,0,0,1,0,0,0,0,0,0,0,0,1,0,0,1,0,0,
%T A037011 1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,1,0,0,0,0,0,1,
%U A037011 0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
%N A037011 Baum-Sweet cubic sequence.
%C A037011 Memo: more sequences like this should be added to the database.
%D A037011 H. Niederreiter and M. Vielhaber, Tree complexity and a doubly ..., J. 
               Complexity, 12 (1996), 187-198.
%H A037011 J.-P. Allouche, <a href="http://www.mat.univie.ac.at/~slc/s/s30allouche.html">
               Finite automata and arithmetic</a> Seminaire Lotharingien de Combinatoire, 
               B30c (1993), 23 pp. [Formerly: Publ. I.R.M.A. Strasbourg, 1993, 1993/
               034, p. 1-18.]
%H A037011 Michael Gilleland, <a href="selfsimilar.html">Some Self-Similar Integer 
               Sequences</a>
%H A037011 D. P. Robbins, <a href="http://arXiv.org/pdf/math.NT/9903092">Cubic Laurent 
               series in characteristic 2 with bounded partial quotients</a>
%F A037011 G.f. satisfies A^3+x^(-1)*A+1 = 0 (mod 2).
%F A037011 It appears that a(n)=sum(k=0, n-1, C(n-1+k, n-1-k)*C(n-1, k)) modulo 
               2 = A082759(n-1) (mod 2). It appears also that a(k)=1 iff k/3 is 
               in A003714. - Benoit Cloitre (benoit7848c(AT)orange.fr), Jun 20 2003
%p A037011 A := x; for n from 1 to 100 do series(x+x*A^3+O(x^(n+2)),x,n+2); A := 
               series(% mod 2,x,n+2); od: A;
%Y A037011 Cf. A086747.
%Y A037011 Sequence in context: A014135 A014054 A014099 this_sequence A024692 A079978 
               A164704
%Y A037011 Adjacent sequences: A037008 A037009 A037010 this_sequence A037012 A037013 
               A037014
%K A037011 nonn,easy
%O A037011 1,1
%A A037011 N. J. A. Sloane (njas(AT)research.att.com).