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Search: id:A085357
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| A085357 |
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Common residues of binomial(3n,n)/(2n+1) modulo 2: relates ternary trees (A001764) to the infinite Fibonacci word (A003849). |
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6
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| 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
(list; graph; listen)
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OFFSET
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0,1
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COMMENT
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The n-th runs of ones is given by: 3 - A003849(n) (infinite Fibonacci word) = A076662(n+1). Runs of zeros are given by: A085358, and are also directly related to the Fibonacci sequence. Coefficients of A(x)^3 are found in A085359.
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LINKS
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J.-P. Allouche, F. v. Haeseler, H.-O. Peitgen and G. Skordev, Linear cellular automata, finite automata and Pascal's triangle, Discrete Appl. Math. 66 (1996), 1-22.
J.-P. Allouche, F. v. Haeseler, H.-O. Peitgen, A. Petersen and G. Skordev, Automaticity of double sequences generated by one-dimensional linear cellular automata, Theoret. Comput. Sci. 186 (1997), 195-209.
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FORMULA
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G.f.: 1 + x*A(x)^3 = A(x) (Mod 2); a(n) = A001764(n) (Mod 2).
a(n)=binomial(3n, n) (mod 2). Characteristic function of Fibbinary numbers (i.e. a(n)=1 iff n is in A003714). - Benoit Cloitre, Nov 15 2003
Recurrence: a(0) = 1, a(2n) = a(4n+1) = a(n), a(4n+3) = 0.
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CROSSREFS
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Cf. A001764 (ternary trees), A085358 (runs of zeros), A076662 (runs of ones), A003849 (infinite Fibonacci word), A085359 (A(x)^3).
Sequence in context: A116938 A105589 A097806 this_sequence A132971 A011748 A130304
Adjacent sequences: A085354 A085355 A085356 this_sequence A085358 A085359 A085360
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KEYWORD
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nonn
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AUTHOR
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Paul D. Hanna (pauldhanna(AT)juno.com), Jun 25 2003
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