A078588 a(n) = 1 if the integer multiple of phi nearest n is greater than n, otherwise 0, where phi = (1+sqrt(5))/2.

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

sqrt(2) A197879 A120243 A120749

sqrt(3) A190669 A190670 A190671

e A190843 A190847 A190860

Pi A191153 A191159 A191164

sqrt(5) A188257 A188258 A188259

sqrt(6) A327253 A327254 A327255

sqrt(8) A327256 A327257 A327258

Guide to related sequences, for k = 3:

** r ********* {a(n)} pos. of 0's pos. of 1's pos. of 2's

(1+sqrt(5))/2 A140397 A140398 A140399 A140400

sqrt(2) A190487 A190488 A190489 A190490

sqrt(3) A190676 A190677 A190678 A190679

e A190893 A191103 A191104 A191105

Pi A327298 A327299 A327300 A327301

sqrt(5) A189463 A189464 A189465 A190158

sqrt(6) A327306 A327307 A327308 A327309

sqrt(8) A327310 A327311 A327312 A327313

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

sqrt(2) A190544 A190545 A190546 A190547 A190548

sqrt(5) A189480 A190813 A190883 A190884 A190885

(End)

References

D. L. Silverman, J. Recr. Math. 9 (4) 208, problem 567 (1976-77).

Links

T. D. Noe, Table of n, a(n) for n = 0..1000

K. Alladi et al., On additive partitions of integers, Discrete Math., 22 (1978), 201-211.

T. Chow, A new characterization of the Fibonacci-free partition, Fibonacci Q. 29 (1991), 174-180; also online here

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

Crossrefs

Cf. A005652, A005653, A024569, A089808, A089809, A140397, A140398, A140399, A140400, A140401.

Sequence in context: A197879 A296657 A082848 * A175992 A173922 A213676

Adjacent sequences: A078585 A078586 A078587 * A078589 A078590 A078591

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