cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 13 results. Next

A140398 Numbers n such that A140397(n) = 0.

Original entry on oeis.org

2, 5, 7, 10, 13, 15, 18, 23, 26, 28, 31, 34, 36, 39, 44, 47, 49, 52, 57, 60, 62, 65, 68, 70, 73, 78, 81, 83, 86, 89, 91, 94, 96, 99, 102, 104, 107, 112, 115, 117, 120, 123, 125, 128, 133, 136, 138, 141, 146, 149, 151, 154, 157, 159, 162, 167, 170, 172, 175, 178, 180
Offset: 1

Views

Author

Fred Lunnon, Jun 20 2008

Keywords

Programs

  • Mathematica
    Select[Range[200],Floor[3*GoldenRatio*#]-3Floor[GoldenRatio*#]==0&] (* Harvey P. Dale, Feb 02 2017 *)

Extensions

Extended by Max Alekseyev, Apr 11 2009

A140399 Numbers n such that A140397(n) = 1.

Original entry on oeis.org

1, 4, 9, 12, 14, 17, 20, 22, 25, 30, 33, 35, 38, 41, 43, 46, 48, 51, 54, 56, 59, 64, 67, 69, 72, 75, 77, 80, 85, 88, 90, 93, 98, 101, 103, 106, 109, 111, 114, 119, 122, 124, 127, 130, 132, 135, 140, 143, 145, 148, 153, 156, 158, 161, 164, 166, 169, 174, 177, 179, 182
Offset: 1

Views

Author

Fred Lunnon, Jun 20 2008

Keywords

Comments

Alternatively, those n for which no block of length 3n in the Fibonacci word A003849 forms an abelian cube (that is, a block of the form x x' x'' where x' and x'' are permutations of x). Proved with the Walnut theorem prover. - Jeffrey Shallit, Jun 23 2021

Extensions

More terms from Max Alekseyev, Apr 24 2010

A140400 Numbers n such that A140397(n) = 2.

Original entry on oeis.org

3, 6, 8, 11, 16, 19, 21, 24, 27, 29, 32, 37, 40, 42, 45, 50, 53, 55, 58, 61, 63, 66, 71, 74, 76, 79, 82, 84, 87, 92, 95, 97, 100, 105, 108, 110, 113, 116, 118, 121, 126, 129, 131, 134, 137, 139, 142, 144, 147, 150, 152, 155, 160, 163, 165, 168, 171, 173, 176
Offset: 1

Views

Author

Fred Lunnon, Jun 20 2008

Keywords

Crossrefs

Extensions

More terms from Rémy Sigrist, Jul 08 2019

A190427 a(n) = [(b*n+c)*r] - b*[n*r] - [c*r], where (r,b,c)=(golden ratio,2,1) and []=floor.

Original entry on oeis.org

1, 1, 2, 1, 0, 2, 1, 2, 1, 0, 2, 1, 0, 1, 1, 2, 1, 0, 2, 1, 2, 1, 1, 2, 1, 0, 1, 1, 2, 1, 0, 2, 1, 0, 1, 1, 2, 1, 0, 2, 1, 2, 1, 1, 2, 1, 0, 1, 1, 2, 1, 0, 2, 1, 2, 1, 1, 2, 1, 0, 2, 1, 2, 1, 0, 2, 1, 0, 1, 1, 2, 1, 0, 2, 1, 2, 1, 1, 2, 1, 0, 1, 1, 2, 1, 0, 2, 1, 0, 1, 1, 2, 1, 0, 2, 1, 2, 1, 0, 2, 1, 0, 1, 1, 2, 1, 0, 2, 1, 2, 1, 1, 2, 1, 0, 2, 1, 2, 1, 0, 2, 1, 0, 1, 1, 2, 1, 0, 2, 1, 2
Offset: 1

Views

Author

Clark Kimberling, May 10 2011

Keywords

Comments

Write a(n) = [(b*n+c)*r] - b*[n*r] - [c*r]. If r>0 and b and c are integers satisfying b>=2 and 0<=c<=b-1, then 0<=a(n)<=b. The positions of 0 in the sequence a are of interest, as are the position sequences for 1,2,...,b. These b+1 position sequences comprise a partition of the positive integers.
Examples:
(golden ratio,2,0): A078588, A005653, A005652
(golden ratio,2,1): A190427 - A190430
(golden ratio,3,0): A140397 - A190400
(golden ratio,3,1): A140431 - A190435
(golden ratio,3,2): A140436 - A190439

Examples

			a(1)=[3r]-2[r]-1=4-3-1=1.
a(2)=[5r]-2[2r]-1=8-6-1=1.
a(3)=[7r]-2[3r]-1=11-8-1=2.
		

Crossrefs

Programs

  • Magma
    [Floor((2*n+1)*(1+Sqrt(5))/2) - 2*Floor(n*(1+Sqrt(5))/2) - 1: n in [1..100]]; // G. C. Greubel, Apr 06 2018
  • Mathematica
    r = GoldenRatio; b = 2; c = 1;
    f[n_] := Floor[(b*n + c)*r] - b*Floor[n*r] - Floor[c*r];
    t = Table[f[n], {n, 1, 320}] (* A190427 *)
    Flatten[Position[t, 0]] (* A190428 *)
    Flatten[Position[t, 1]] (* A190429 *)
    Flatten[Position[t, 2]] (* A190430 *)
    Table[Floor[(2*n+1)*GoldenRatio] - 2*Floor[n*GoldenRatio] -1, {n,1,100}] (* G. C. Greubel, Apr 06 2018 *)
  • PARI
    for(n=1,100, print1(floor((2*n+1)*(1+sqrt(5))/2) - 2*floor(n*(1+sqrt(5))/2) - 1, ", ")) \\ G. C. Greubel, Apr 06 2018
    
  • Python
    from mpmath import mp, phi
    from sympy import floor
    mp.dps=100
    def a(n): return floor((2*n + 1)*phi) - 2*floor(n*phi) - 1
    print([a(n) for n in range(1, 132)]) # Indranil Ghosh, Jul 02 2017
    

Formula

a(n) = [(2*n+1)*r] - 2*[n*r] - 1, where r=(1+sqrt(5))/2.

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

Original entry on oeis.org

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

Views

Author

Robert G. Wilson v, Dec 02 2002

Keywords

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).

Crossrefs

Programs

  • 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 *)
  • 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

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

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

A190440 [(bn+c)r]-b[nr]-[cr], where (r,b,c)=(golden ratio,4,0) and []=floor.

Original entry on oeis.org

2, 0, 3, 1, 0, 2, 1, 3, 2, 0, 3, 1, 0, 2, 1, 3, 2, 0, 2, 1, 3, 2, 0, 3, 1, 0, 2, 1, 3, 2, 0, 3, 1, 0, 2, 0, 3, 1, 0, 2, 1, 3, 2, 0, 3, 1, 0, 2, 1, 3, 2, 0, 3, 1, 3, 2, 0, 3, 1, 0, 2, 1, 3, 2, 0, 3, 1, 0, 2, 1, 3, 1, 0, 2, 1, 3, 2, 0, 3, 1, 0, 2, 1, 3, 2, 0, 3, 1, 0, 2, 0, 3, 1, 0, 2, 1, 3, 2, 0, 3, 1, 0, 2, 1, 3, 2, 0, 2, 1, 3, 2, 0
Offset: 1

Views

Author

Clark Kimberling, May 10 2011

Keywords

Comments

Write a(n)=[(bn+c)r]-b[nr]-[cr]. If r>0 and b and c are integers satisfying b>=2 and 0<=c<=b-1, then 0<=a(n)<=b. The positions of 0 in the sequence a are of interest, as are the position sequences for 1,2,...,b. These b+1 position sequences comprise a partition of the positive integers.
Examples:
(golden ratio,2,0): A078588, A005653, A005652
(golden ratio,2,1): A190427-A190430
(golden ratio,3,0): A140397-A190400
(golden ratio,3,1): A140431-A190435
(golden ratio,3,2): A140436-A190439

Crossrefs

Programs

  • Mathematica
    r = GoldenRatio;
    f[n_] := Floor[4*n*r] - 4*Floor[n*r];
    t = Table[f[n], {n, 1, 320}] (* A190440 *)
    Flatten[Position[t, 0]]  (* A190240 *)
    Flatten[Position[t, 1]]  (* A190249 *)
    Flatten[Position[t, 2]]  (* A190442 *)
    Flatten[Position[t, 3]]  (* A190443 *)
    Flatten[Position[t, 4]]  (* A190248 *)

Formula

a(n)=[4nr]-4[nr], where r=golden ratio.

A190431 a(n) = [(b*n+c)*r] - b*[n*r] - [c*r], where (r,b,c)=(golden ratio,3,1) and []=floor.

Original entry on oeis.org

2, 1, 3, 2, 0, 2, 1, 3, 2, 1, 3, 1, 0, 2, 1, 3, 2, 0, 2, 1, 3, 2, 1, 3, 1, 0, 2, 1, 3, 2, 1, 2, 1, 0, 2, 1, 3, 2, 0, 2, 1, 3, 2, 1, 3, 1, 0, 2, 1, 3, 2, 1, 2, 1, 3, 2, 1, 3, 2, 0, 2, 1, 3, 2, 1, 2, 1, 0, 2, 1, 3, 2, 0, 2, 1, 3, 2, 1, 3, 1, 0, 2, 1, 3, 2, 1, 2, 1, 0, 2, 1, 3, 2, 0, 2, 1, 3, 2, 1, 3, 1, 0, 2, 1, 3, 2, 1, 2, 1, 3, 2, 1, 3, 1, 0, 2, 1, 3, 2, 1, 2, 1, 0, 2, 1, 3, 2, 0, 2, 1, 3, 2, 1, 3, 1, 0
Offset: 1

Views

Author

Clark Kimberling, May 10 2011

Keywords

Comments

Write a(n) = [(b*n+c)*r] - b*[n*r] - [c*r]. If r>0 and b and c are integers satisfying b>=2 and 0<=c<=b-1, then 0<=a(n)<=b. The positions of 0 in the sequence a are of interest, as are the position sequences for 1,2,...,b. These b+1 position sequences comprise a partition of the positive integers.
Examples:
(golden ratio,2,0): A078588, A005653, A005652
(golden ratio,2,1): A190427 - A190430
(golden ratio,3,0): A140397 - A190400
(golden ratio,3,1): A140431 - A190435
(golden ratio,3,2): A140436 - A190439

Crossrefs

Programs

  • Magma
    [Floor((3*n+1)*(1+Sqrt(5))/2) - 3*Floor(n*(1+Sqrt(5))/2) - 1: n in [1..100]]; // G. C. Greubel, Apr 06 2018
  • Mathematica
    r = GoldenRatio; b = 3; c = 1;
    f[n_] := Floor[(b*n + c)*r] - b*Floor[n*r] - Floor[c*r];
    t = Table[f[n], {n, 1, 320}] (* A190431 *)
    Flatten[Position[t, 0]] (* A190432 *)
    Flatten[Position[t, 1]] (* A190433 *)
    Flatten[Position[t, 2]] (* A190434 *)
    Flatten[Position[t, 3]] (* A190435 *)
  • PARI
    for(n=1,100, print1(floor((3*n+1)*(1+sqrt(5))/2) - 3*floor(n*(1+sqrt(5))/2) - 1, ", ")) \\ G. C. Greubel, Apr 06 2018
    

Formula

a(n) = floor((3*n+1)*(1+sqrt(5))/2) - 3*floor(n*(1+sqrt(5))/2) - 1. - G. C. Greubel, Apr 06 2018

A190436 a(n) = [(b*n+c)*r] - b*[n*r] - [c*r], where (r,b,c)=(golden ratio,3,2) and []=floor.

Original entry on oeis.org

2, 0, 2, 1, 0, 2, 1, 3, 1, 0, 2, 1, 0, 2, 1, 2, 1, 0, 2, 1, 3, 2, 0, 2, 1, 0, 2, 1, 3, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 3, 1, 0, 2, 1, 0, 2, 1, 2, 1, 0, 2, 1, 3, 2, 0, 2, 1, 0, 2, 1, 3, 1, 0, 2, 1, 0, 2, 1, 2, 1, 0, 2, 1, 3, 2, 0, 2, 1, 0, 2, 1, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 3, 1, 0, 2, 1, 0, 2, 1, 2, 1, 0, 2, 1, 3, 2, 0, 2, 1, 0
Offset: 1

Views

Author

Clark Kimberling, May 10 2011

Keywords

Comments

Write a(n)=[(bn+c)r]-b[nr]-[cr]. If r>0 and b and c are integers satisfying b>=2 and 0<=c<=b-1, then 0<=a(n)<=b. The positions of 0 in the sequence a are of interest, as are the position sequences for 1,2,...,b. These b+1 position sequences comprise a partition of the positive integers.
Examples:
(golden ratio,2,0): A078588, A005653, A005652
(golden ratio,2,1): A190427 - A190430
(golden ratio,3,0): A140397 - A190400
(golden ratio,3,1): A140431 - A190435
(golden ratio,3,2): A140436 - A190439
(golden ratio,4,c): A140440 - A190461

Crossrefs

Programs

  • Mathematica
    r = GoldenRatio; b = 3; c = 2;
    f[n_] := Floor[(b*n + c)*r] - b*Floor[n*r] - Floor[c*r];
    t = Table[f[n], {n, 1, 320}]
    Flatten[Position[t, 0]] (* A190437 *)
    Flatten[Position[t, 1]] (* A190438 *)
    Flatten[Position[t, 2]] (* A190439 *)
    Flatten[Position[t, 3]] (* A302253 *)

A190445 [(bn+c)r]-b[nr]-[cr], where (r,b,c)=(golden ratio,4,1) and []=floor.

Original entry on oeis.org

3, 1, 4, 2, 0, 3, 1, 4, 2, 1, 3, 2, 0, 3, 1, 4, 2, 1, 3, 2, 4, 3, 1, 3, 2, 0, 3, 1, 4, 2, 1, 3, 2, 0, 3, 1, 4, 2, 1, 3, 1, 4, 2, 1, 3, 2, 0, 3, 1, 4, 2, 1, 3, 2, 4, 3, 1, 4, 2, 0, 3, 1, 4, 2, 1, 3, 2, 0, 3, 1, 4, 2, 1, 3, 2, 4, 2, 1, 3, 2, 0, 3, 1, 4, 2, 1, 3, 2, 0, 3, 1, 4, 2, 0, 3, 1, 4, 2, 1, 3, 2, 0, 3, 1, 4, 2, 1, 3, 2, 4, 3, 1, 3, 2, 0, 3, 1, 4, 2, 1, 3, 2, 0, 3, 1, 4, 2, 1, 3, 1, 4, 2, 1, 3, 2, 0
Offset: 1

Views

Author

Clark Kimberling, May 10 2011

Keywords

Comments

Write a(n)=[(bn+c)r]-b[nr]-[cr]. If r>0 and b and c are integers satisfying b>=2 and 0<=c<=b-1, then 0<=a(n)<=b. The positions of 0 in the sequence a are of interest, as are the position sequences for 1,2,...,b. These b+1 position sequences comprise a partition of the positive integers.
Examples:
(golden ratio,2,0): A078588, A005653, A005652
(golden ratio,2,1): A190427-A190430
(golden ratio,3,0): A140397-A190400
(golden ratio,3,1): A140431-A190435
(golden ratio,3,2): A140436-A190439
(golden ratio,4,c): A190440-A190461

Crossrefs

Programs

  • Mathematica
    r = GoldenRatio; b = 4; c = 1;
    f[n_] := Floor[(b*n + c)*r] - b*Floor[n*r] - Floor[c*r];
    t = Table[f[n], {n, 1, 320}]
    Flatten[Position[t, 0]]
    Flatten[Position[t, 1]]
    Flatten[Position[t, 2]]
    Flatten[Position[t, 3]]
    Flatten[Position[t, 4]]

A190457 a(n) = [(bn+c)r]-b[nr]-[cr], where (r,b,c)=(golden ratio,4,3) and []=floor.

Original entry on oeis.org

3, 1, 4, 2, 1, 3, 2, 4, 3, 1, 4, 2, 0, 3, 1, 4, 2, 1, 3, 2, 4, 3, 1, 4, 2, 1, 3, 2, 4, 3, 1, 3, 2, 0, 3, 1, 4, 2, 1, 3, 2, 4, 3, 1, 4, 2, 1, 3, 1, 4, 2, 1, 3, 2, 4, 3, 1, 4, 2, 1, 3, 2, 4, 3, 1, 4, 2, 0, 3, 1, 4, 2, 1, 3, 2, 4, 3, 1, 4, 2, 1, 3, 2, 4, 2, 1, 3, 2, 0, 3, 1, 4, 2, 1, 3, 2, 4, 3, 1, 4, 2, 1, 3, 1, 4, 2, 1, 3, 2, 4, 3, 1, 4, 2, 1, 3, 2, 4, 3, 1, 3, 2, 0, 3, 1, 4, 2, 1, 3, 2, 4, 3
Offset: 1

Views

Author

Clark Kimberling, May 10 2011

Keywords

Comments

Write a(n)=[(bn+c)r]-b[nr]-[cr]. If r>0 and b and c are integers satisfying b>=2 and 0<=c<=b-1, then 0<=a(n)<=b. The positions of 0 in the sequence a are of interest, as are the position sequences for 1,2,...,b. These b+1 position sequences comprise a partition of the positive integers.
Examples:
(golden ratio,2,0): A078588, A005653, A005652
(golden ratio,2,1): A190427-A190430
(golden ratio,3,0): A140397-A190400
(golden ratio,3,1): A140431-A190435
(golden ratio,3,2): A140436-A190439
(golden ratio,4,c): A190440-A190461

Crossrefs

Programs

  • Mathematica
    r = GoldenRatio; b = 4; c = 3;
    f[n_] := Floor[(b*n + c)*r] - b*Floor[n*r] - Floor[c*r];
    t = Table[f[n], {n, 1, 320}]
    Flatten[Position[t, 0]]
    Flatten[Position[t, 1]]
    Flatten[Position[t, 2]]
    Flatten[Position[t, 3]]
    Flatten[Position[t, 4]]
Showing 1-10 of 13 results. Next