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A037011 Baum-Sweet cubic sequence. 13
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, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 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.]
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
Sequence in context: A014135 A014054 A014099 * A070563 A024692 A079978
KEYWORD
nonn,easy
AUTHOR
STATUS
approved

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Last modified March 28 09:04 EDT 2024. Contains 371240 sequences. (Running on oeis4.)