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 A037011 Baum-Sweet cubic sequence. 12
 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, 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, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Memo: more sequences like this should be added to the database. LINKS J.-P. Allouche, Finite automata and arithmetic Seminaire Lotharingien de Combinatoire, B30c (1993), 23 pp. [Formerly: Publ. I.R.M.A. Strasbourg, 1993, 1993/034, p. 1-18.] Michael Gilleland, Some Self-Similar Integer Sequences H. Niederreiter and M. Vielhaber, Tree complexity and a doubly exponential gap between structured and random sequences, J. Complexity, 12 (1996), 187-198. D. P. Robbins, Cubic Laurent series in characteristic 2 with bounded partial quotients, arXiv:math/9903092 [math.NT], 1999. FORMULA G.f. satisfies A^3+x^(-1)*A+1 = 0 (mod 2). 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 From Antti Karttunen, Nov 03 2017: (Start) If Cloitre's above observation holds, then we also have (assuming starting offset 0, with a(0) = 1): a(n) = A000035(A106737(n)) a(n) = A010052(A005940(1+n)). (End) MAPLE 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; MATHEMATICA m = 100; A[_] = 0; Do[A[x_] = x + x A[x]^3 + O[x]^m // Normal // PolynomialMod[#, 2]&, {m}]; CoefficientList[A[x], x] // Rest (* Jean-François Alcover, Oct 15 2019 *) CROSSREFS Cf. A086747, A106737, A277335. Sequence in context: A014135 A014054 A014099 * A070563 A024692 A079978 Adjacent sequences:  A037008 A037009 A037010 * A037012 A037013 A037014 KEYWORD nonn,easy AUTHOR STATUS approved

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Last modified April 16 07:59 EDT 2021. Contains 343030 sequences. (Running on oeis4.)