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A078588
a(n) = 1 if the integer multiple of phi nearest n is greater than n, otherwise 0, where phi = (1+sqrt(5))/2.
28
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
OFFSET
0,1
COMMENTS
From Fred Lunnon, Jun 20 2008: (Start)
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.
This is the complement of A089809; also a(n) = 1 iff A024569(n) = 1. - Gary W. Adamson, Nov 11 2003
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
From Clark Kimberling and Jianing Song, Sep 09 2019: (Start)
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
(1+sqrt(5))/2 A078588 A005653 A005652
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)
REFERENCES
D. L. Silverman, J. Recr. Math. 9 (4) 208, problem 567 (1976-77).
LINKS
K. Alladi et al., On additive partitions of integers, Discrete Math., 22 (1978), 201-211.
T. Y. Chow and C. D. Long, Additive partitions and continued fractions, Ramanujan J., 3 (1999), 55-72 [set alpha=(1+sqrt(5))/2 in Theorem 2 to get A005652 and A005653].
FORMULA
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) = round(phi*n) - floor(phi*n). - Michel Dekking, Sep 30 2016
a(n) = (n+floor(n*sqrt(5))) mod 2. - Chai Wah Wu, Aug 17 2022
MATHEMATICA
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}]
(* Clark Kimberling, Aug 26 2019 *)
PROG
(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
def A078588(n): return (n+isqrt(5*n**2))&1 # Chai Wah Wu, Aug 17 2022
KEYWORD
easy,nonn
AUTHOR
Robert G. Wilson v, Dec 02 2002
EXTENSIONS
Edited by N. J. A. Sloane, Jun 20 2008, at the suggestion of Fred Lunnon
Edited by Jianing Song, Sep 09 2019
Offset corrected by Jianing Song, Sep 10 2019
STATUS
approved