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%I #29 Jun 24 2023 22:10:57
%S 1,0,0,1,0,0,1,0,0,0,0,0,1,0,0,1,0,0,0,0,0,0,0,0,1,0,0,1,0,0,1,0,0,0,
%T 0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,1,0,0,0,0,0,1,0,0,1,0,0,0,0,
%U 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,1,0,0
%N Baum-Sweet cubic sequence.
%C Memo: more sequences like this should be added to the database.
%H 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 Michael Gilleland, <a href="/selfsimilar.html">Some Self-Similar Integer Sequences</a>
%H H. Niederreiter and M. Vielhaber, <a href="http://dx.doi.org/10.1006/jcom.1996.0014">Tree complexity and a doubly exponential gap between structured and random sequences</a>, J. Complexity, 12 (1996), 187-198.
%H D. P. Robbins, <a href="https://arxiv.org/abs/math/9903092">Cubic Laurent series in characteristic 2 with bounded partial quotients</a>, arXiv:math/9903092 [math.NT], 1999.
%H <a href="/index/Ch#char_fns">Index entries for characteristic functions</a>
%F G.f. satisfies A^3+x^(-1)*A+1 = 0 (mod 2).
%F 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_, Jun 20 2003
%F From _Antti Karttunen_, Nov 03 2017: (Start)
%F If Cloitre's above observation holds, then we also have (assuming starting offset 0, with a(0) = 1):
%F a(n) = A000035(A106737(n))
%F a(n) = A010052(A005940(1+n)).
%F (End)
%p 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;
%t m = 100; A[_] = 0;
%t Do[A[x_] = x + x A[x]^3 + O[x]^m // Normal // PolynomialMod[#, 2]&, {m}];
%t CoefficientList[A[x], x] // Rest (* _Jean-François Alcover_, Oct 15 2019 *)
%Y Cf. A086747, A106737, A277335.
%K nonn,easy
%O 1,1
%A _N. J. A. Sloane_.
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