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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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16
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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
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OFFSET
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0,1
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COMMENTS
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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.
a(n) = 0 iff some binary digit of n is 1 while the corresponding binary digit of 3*n is 0. - Robert Israel, Jul 12 2016
The Run Length Transform of [0,1,0,0,0,...], A063524, the characteristic function of 1. (See A227349 for the definition). - Antti Karttunen, Oct 15 2016
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LINKS
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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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MAPLE
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f:= proc(n) local L, Lp;
L:= convert(n, base, 2);
Lp:= convert(3*n, base, 2);
if has(L-Lp[1..nops(L)], 1) then 0 else 1 fi
end proc:
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MATHEMATICA
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PROG
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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