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A078588
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a(n) = 1 if the integer multiple of phi nearest n is greater than n, otherwise 0, where phi = (1+sqrt(5))/2.
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28
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0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1
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OFFSET
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0,1
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COMMENTS
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Partition the positive integers into two sets A_0 and A_1 defined by A_k == { n | a(n) = k }; so A_0 = A005653 = { 2, 4, 5, 7, 10, 12, 13, 15, 18, 20, ... }, A_1 = A005652 = { 1, 3, 6, 8, 9, 11, 14, 16, 17, 19, 21, ... }.
Then form the sets of sums of pairs of distinct elements from each set and take the complement of their union: this is the Fibonacci numbers { 1, 2, 3, 5, 8, 13, 21, 34, 55, ... } (see the Chow article). (End)
The Chow-Long paper gives a connection with continued fractions, as well as generalizations and other references for this and related sequences.
Since (n*phi) is equidistributed, s(n):=(Sum_{k=1..n}a(k))/n converges to 1/2, but actually s(n) is exactly equal to 1/2 for many values of n. These values are given by A194402. - Michel Dekking, Sep 30 2016
Suppose that k >= 2, and let a(n) = floor(n*k*r) - k*floor(n*r) = k*{n*r} - {n*k*r}, an integer strictly between 0 and k, where {} denotes fractional part. For h = 0,1,...,k-1, let s(h) be the sequence of positions of h in {a(n)}. The sets s(h) partition the positive integers. Although a(n)/n -> k, the sequence a(n)-k*n appears to be unbounded.
Guide to related sequences, for k = 2:
** r ********* {a(n)} positions of 0's positions of 1's
Guide to related sequences, for k = 3:
** r ********* {a(n)} pos. of 0's pos. of 1's pos. of 2's
Guide to related sequences, for k = 4:
** r ********* {a(n)} pos. of 0's pos. of 1's pos. of 2's pos. of 3's
(End)
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REFERENCES
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D. L. Silverman, J. Recr. Math. 9 (4) 208, problem 567 (1976-77).
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LINKS
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FORMULA
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a(n) = floor(2*phi*n) - 2*floor(phi*n) where phi denotes the golden ratio (1 + sqrt(5))/2. - Fred Lunnon, Jun 20 2008
a(n) = 2{n*phi} - {2n*phi}, where { } denotes fractional part. - Clark Kimberling, Jan 01 2007
a(n) = n + 1 + ceiling(n*sqrt(5)) - 2*ceiling(n*phi) where phi = (1+sqrt(5))/2. - Benoit Cloitre, Dec 05 2002
a(n) = (n+floor(n*sqrt(5))) mod 2. - Chai Wah Wu, Aug 17 2022
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MATHEMATICA
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f[n_] := Block[{k = Floor[n/GoldenRatio]}, If[n - k*GoldenRatio > (k + 1)*GoldenRatio - n, 1, 0]]; Table[ f[n], {n, 0, 105}]
r = (1 + Sqrt[5])/2; z = 300;
t = Table[Floor[2 n*r] - 2 Floor[n*r], {n, 0, z}]
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PROG
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(PARI) a(n)=if(n, n+1+ceil(n*sqrt(5))-2*ceil(n*(1+sqrt(5))/2), 0) \\ (changed by Jianing Song, Sep 10 2019 to include a(0) = 0)
(Python)
from math import isqrt
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CROSSREFS
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Cf. A001622, A005652, A005653, A024569, A089808, A089809, A140397, A140398, A140399, A140400, A140401.
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KEYWORD
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easy,nonn,changed
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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