A327313 -id:A327313 - cp's OEIS frontend

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

Robert G. Wilson v, Dec 02 2002

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)

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

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].

Cf. A001622, A005652, A005653, A024569, A089808, A089809, A140397, A140398, A140399, A140400, A140401.

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 *)

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)

from math import isqrt
def A078588(n): return (n+isqrt(5*n**2))&1 # Chai Wah Wu, Aug 17 2022

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

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

A327310 a(n) = floor(3nr) - 3floor(nr), where r = sqrt(8).

Original entry on oeis.org

0, 2, 1, 1, 0, 0, 2, 2, 1, 1, 0, 0, 2, 2, 1, 1, 0, 0, 2, 2, 1, 1, 0, 0, 2, 2, 1, 1, 0, 0, 2, 2, 1, 1, 0, 2, 2, 1, 1, 0, 0, 2, 2, 1, 1, 0, 0, 2, 2, 1, 1, 0, 0, 2, 2, 1, 1, 0, 0, 2, 2, 1, 1, 0, 0, 2, 2, 1, 0, 0, 2, 2, 1, 1, 0, 0, 2, 2, 1, 1, 0, 0, 2, 2, 1, 1

Offset: 0

Clark Kimberling, Sep 07 2019

Clark Kimberling, Table of n, a(n) for n = 0..10000

The positions of 0's, 1's and 2's in {a(n) : n > 0} are given by A327311, A327312 and A327313.

r = Sqrt[8]; z = 300;
t = Table[Floor[3 n*r] - 3 Floor[n*r], {n, 0, z}]

a(n) = floor(3*n*r) - 3*floor(n*r), where r = sqrt(8).

A327311 Positions of 0's in {A327310(n) : n > 0}.

Original entry on oeis.org

4, 5, 10, 11, 16, 17, 22, 23, 28, 29, 34, 39, 40, 45, 46, 51, 52, 57, 58, 63, 64, 68, 69, 74, 75, 80, 81, 86, 87, 92, 93, 98, 99, 103, 104, 109, 110, 115, 116, 121, 122, 127, 128, 133, 134, 138, 139, 144, 145, 150, 151, 156, 157, 162, 163, 168, 169, 173, 174

Offset: 1

Clark Kimberling, Sep 07 2019

The positive integers are partitioned by A327311, A327312, and A327313.
Although a(n)/n->3, the sequence a(n)-3n appears to be unbounded.
Positive integers k such that A327310(k) = 0. - Jianing Song, Sep 30 2019

Clark Kimberling and Jianing Song, Table of n, a(n) for n = 1..10000

Cf. A327310, A327312, A327313.

r = Sqrt[8]; z = 300;
t = Table[Floor[3 n*r] - 3 Floor[n*r], {n, 1, z}]  (* {A327310(n) : n > 0} *)
Flatten[Position[t, 0]]  (* A327311 *)
Flatten[Position[t, 1]]  (* A327312 *)
Flatten[Position[t, 2]]  (* A327313 *)

Corrected by Jianing Song, Sep 30 2019

A327312 Positions of 1's in {A327310(n) : n > 0}.

Original entry on oeis.org

2, 3, 8, 9, 14, 15, 20, 21, 26, 27, 32, 33, 37, 38, 43, 44, 49, 50, 55, 56, 61, 62, 67, 72, 73, 78, 79, 84, 85, 90, 91, 96, 97, 102, 107, 108, 113, 114, 119, 120, 125, 126, 131, 132, 136, 137, 142, 143, 148, 149, 154, 155, 160, 161, 166, 167, 171, 172

Offset: 1

Clark Kimberling, Sep 07 2019

The positive integers are partitioned by A327311, A327312, and A327313.
Although a(n)/n->3, the sequence a(n)-3n appears to be unbounded.
Positive integers k such that A327310(k) = 1. - Jianing Song, Sep 30 2019

Clark Kimberling and Jianing Song, Table of n, a(n) for n = 1..10000

Cf. A327310, A327311, A327313.

r = Sqrt[8]; z = 300;
t = Table[Floor[3 n*r] - 3 Floor[n*r], {n, 1, z}]  (* {A327310(n) : n > 0} *)
Flatten[Position[t, 0]]  (* A327311 *)
Flatten[Position[t, 1]]  (* A327312 *)
Flatten[Position[t, 2]]  (* A327313 *)

Corrected by Jianing Song, Sep 30 2019

cp's OEIS Frontend

A078588 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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A327310 a(n) = floor(3*n*r) - 3*floor(n*r), where r = sqrt(8).

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A327311 Positions of 0's in {A327310(n) : n > 0}.

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Programs

Mathematica

Extensions

A327312 Positions of 1's in {A327310(n) : n > 0}.

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Author

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Programs

Mathematica

Extensions

A327310 a(n) = floor(3nr) - 3floor(nr), where r = sqrt(8).