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 19 results. Next

A329825 Beatty sequence for (3+sqrt(17))/4.

Original entry on oeis.org

1, 3, 5, 7, 8, 10, 12, 14, 16, 17, 19, 21, 23, 24, 26, 28, 30, 32, 33, 35, 37, 39, 40, 42, 44, 46, 48, 49, 51, 53, 55, 56, 58, 60, 62, 64, 65, 67, 69, 71, 73, 74, 76, 78, 80, 81, 83, 85, 87, 89, 90, 92, 94, 96, 97, 99, 101, 103, 105, 106, 108, 110, 112, 113
Offset: 1

Views

Author

Clark Kimberling, Nov 22 2019

Keywords

Comments

Let r = (3+sqrt(17))/4. Then (floor(n*r)) and (floor(n*r + r/2)) are a pair of Beatty sequences; i.e., every positive integer is in exactly one of the sequences. The sequence (a(n) mod 2) of 0's and 1's has only two run-lengths: 4 and 5.
More generally, suppose that t > 0. There exists an irrational number r such that (floor(n*r)) and (floor(n*(r+t))) are a pair of Beatty sequences. Specifically, r = (2 - t + sqrt(t^2 + 4))/2, as in the Mathematica code below. See Comments at A182760.
************
Guide to related sequences:
t = 1: A000201 and A001950 (Wythoff sequences), r = (1+sqrt(5))/2
t = 1/2: A329825 and A329826, r = (3 + sqrt(17))/4
t = 1/3: A329827 and A329828, r = (5 + sqrt(37))/6
t = 2/3: A329829 and A329830, r = (2 + sqrt(10))/3
t = 1/4: A329831 and A329832, r = (7 + sqrt(65))/8
t = 3/4: A329833 and A329834, r = (5 + sqrt(73))/8
t = 1/5: A329835 and A329836, r = (9 + sqrt(101))/10
t = 2/5: A329837 and A329838, r = (4 + sqrt(26))/5
t = 5/2: A329839 and A329840, r = (-1 + sqrt(41))/4
t = 3/5: A329841 and A329842, r = (7 + sqrt(109))/10
t = 5/3: A329843 and A329844, r = (1 + sqrt(61))/6
t = 5/4: A329847 and A329848, r = (3 + sqrt(89))/8
t = 4/5: A329845 and A329846, r = (3 + sqrt(29))/5
t = 6/5: A329923 and A329924, r = (2 + sqrt(34))/5
t = 8/5: A329925 and A329926, r = (1 + sqrt(41))/5
t = 2: A001951 and A001952, r = sqrt(2)
t = 3: A001961 and A004976, r = -1 + sqrt(5)
t = 4: A001961 and A001962, r = -1 + sqrt(5)
t = 5: A184522 and A184523, r = (-3 + sqrt(29))/2
t = 6: A187396 and A187395, r = -2 + sqrt(10).
Starts to deviate from A059565 at a(73). - R. J. Mathar, Nov 26 2019
Sequences for t = 5/4, 4/5 and 3 corrected by Georg Fischer, Aug 22 2021

Links

Crossrefs

Cf. A188485, A329826 (complement), A182760.

Programs

  • Mathematica
    t = 1/2; r = Simplify[(2 - t + Sqrt[t^2 + 4])/2]; s = Simplify[r/(r - 1)];
Table[Floor[r*n], {n, 1, 200}]  (* A329825 *)
Table[Floor[s*n], {n, 1, 200}]  (* A329826 *)
  • PARI
    a(n)=(sqrtint(17*n^2)+3*n)\4 \\ Charles R Greathouse IV, Jan 25 2022

Formula

a(n) = floor(r*n), where r = (3+sqrt(17))/4.

A182761 Beatty sequence for (7 - sqrt(5))/2.

Original entry on oeis.org

2, 4, 7, 9, 11, 14, 16, 19, 21, 23, 26, 28, 30, 33, 35, 38, 40, 42, 45, 47, 50, 52, 54, 57, 59, 61, 64, 66, 69, 71, 73, 76, 78, 80, 83, 85, 88, 90, 92, 95, 97, 100, 102, 104, 107, 109, 111, 114, 116, 119, 121, 123, 126, 128, 131, 133
Offset: 1

Views

Author

Clark Kimberling, Nov 28 2010

Keywords

Comments

See the first comment at A182760; its complement, A182761, gives the positions of the numbers k*sqrt(5) when these numbers and the numbers j*u, where u = golden ratio, are jointly ranked.

Links

Crossrefs

Cf. A182760.

Programs

  • Magma
    [Floor(n*(7-5^(1/2))/2): n in [1..80]]; // Vincenzo Librandi, Oct 25 2011
  • Mathematica
    Table[Floor[n*(7-Sqrt[5])/2], {n,1,80}] (* G. C. Greubel, Jul 08 2018 *)
  • PARI
    for(n=1,80, print1(floor(n*(7-sqrt(5))/2), ", ")) \\ G. C. Greubel, Jul 08 2018

Formula

a(n) = floor(n*(7 - sqrt(5))/2).

Extensions

Typo in formula corrected by Vincenzo Librandi, Oct 25 2011

A182769 Beatty sequence for (4 + sqrt(2))/2.

Original entry on oeis.org

2, 5, 8, 10, 13, 16, 18, 21, 24, 27, 29, 32, 35, 37, 40, 43, 46, 48, 51, 54, 56, 59, 62, 64, 67, 70, 73, 75, 78, 81, 83, 86, 89, 92, 94, 97, 100, 102, 105, 108, 110, 113, 116, 119, 121, 124, 127, 129, 132, 135, 138, 140, 143, 146, 148, 151, 154, 157, 159, 162, 165, 167, 170, 173, 175
Offset: 1

Views

Author

Clark Kimberling, Nov 30 2010

Keywords

Comments

Let u=1+sqrt(2) and v=sqrt(2). Jointly rank {j*u} and {k*v} as in the first comment at A182760; a(n) is the position of n*u.
Is this a shifted version of A126281? - R. J. Mathar, Jan 24 2011
The answer to R. J. Mathar's question is no: A126281 contains 65 while this sequence does not. - L. Edson Jeffery, Sep 02 2014

Links

Crossrefs

Cf. A182760, A182770.

Programs

  • Magma
    [Floor(n*(4 + Sqrt(2))/2): n in [1..50]]; // G. C. Greubel, Jan 27 2018
  • Mathematica
    Table[Floor[n*(4 + Sqrt[2])/2], {n, 1, 100}] (* G. C. Greubel, Jan 27 2018 *)
  • PARI
    a(n) = floor(n*(4+sqrt(2))/2); \\ Michel Marcus, Sep 02 2014

Formula

a(n) = floor(n*(4 + sqrt(2))/2).

A182773 Beatty sequence for 1+2^(2/3).

Original entry on oeis.org

2, 5, 7, 10, 12, 15, 18, 20, 23, 25, 28, 31, 33, 36, 38, 41, 43, 46, 49, 51, 54, 56, 59, 62, 64, 67, 69, 72, 75, 77, 80, 82, 85, 87, 90, 93, 95, 98, 100, 103, 106, 108, 111, 113, 116, 119, 121, 124, 126, 129, 131, 134, 137, 139, 142
Offset: 1

Views

Author

Clark Kimberling, Nov 30 2010

Keywords

Comments

Let u=2^(1/3). Jointly rank {j*u} and {k/u} as in the first comment at A182760; a(n) is the position of n*u.

Links

Crossrefs

Cf. A005480, A182760, A182774.

Programs

  • Mathematica
    Floor[Range[100]*(1 + 2^(2/3))] (* Paolo Xausa, Jul 09 2024 *)

Formula

a(n) = floor(n*(1+2^(2/3))).

A182778 Beatty sequence for 3 + sqrt(3).

Original entry on oeis.org

4, 9, 14, 18, 23, 28, 33, 37, 42, 47, 52, 56, 61, 66, 70, 75, 80, 85, 89, 94, 99, 104, 108, 113, 118, 123, 127, 132, 137, 141, 146, 151, 156, 160, 165, 170, 175, 179, 184, 189, 194, 198, 203, 208, 212, 217, 222, 227, 231, 236, 241, 246, 250, 255
Offset: 1

Views

Author

Clark Kimberling, Nov 30 2010

Keywords

Comments

Let u=2-sqrt(3) and v=1. Jointly rank {ju} and {kv} as in the first comment at A182760; a(n) is the position of n. A182778 is the complement of A182777.

Links

Crossrefs

Cf. A182760, A182777.

Programs

  • Magma
    [Floor(n*(3+Sqrt(3))): n in [1..80]]; // Vincenzo Librandi, Oct 25 2011
  • Mathematica
    Table[Floor[(3+Sqrt[3])*n], {n, 54}]

Formula

a(n) = floor(n*(3 + sqrt(3))).
From Miko Labalan, Dec 17 2016: (Start)
a(n) = 3n + A022838(n);
For n > 0, a(n) = 5*floor(n*(sqrt(3)-1)) + 4*floor(n*(2-sqrt(3))) + 4;
a(0) = 0, a(n) = a(n - 1) + A022838(n) - A022838(n - 1) + 3.
(End)

Extensions

Typo in formula corrected by Vincenzo Librandi, Oct 25 2011

A184808 n + floor(r*n), where r = sqrt(2/3); complement of A184809.

Original entry on oeis.org

1, 3, 5, 7, 9, 10, 12, 14, 16, 18, 19, 21, 23, 25, 27, 29, 30, 32, 34, 36, 38, 39, 41, 43, 45, 47, 49, 50, 52, 54, 56, 58, 59, 61, 63, 65, 67, 69, 70, 72, 74, 76, 78, 79, 81, 83, 85, 87, 89, 90, 92, 94, 96, 98, 99, 101, 103, 105, 107, 108, 110, 112, 114, 116
Offset: 1

Views

Author

Clark Kimberling, Jan 22 2011

Keywords

Comments

This is the Beatty sequence for 1 + sqrt(2/3).
Also, a(n) is the position of 2*n^2 in the sequence obtained by arranging all the numbers in the sets {2*h^2, h >= 1} and {3*k^2, k >= 1} in increasing order. - Clark Kimberling, Oct 20 2014
Also, numbers n such that floor((n+1)*sqrt(6)) - floor(n*sqrt(6)) = 2. - Clark Kimberling, Jul 15 2015

Links

Crossrefs

Cf. A184809, A182760 (comment about joint ranking),
A184810, A184811, A249096.

Programs

  • Magma
    [n+Floor(n*Sqrt(2/3)): n in [1..70]]; // Vincenzo Librandi, Oct 23 2014
  • Mathematica
    r=(2/3)^(1/2); s=(3/2)^(1/2);
a[n_]:=n+Floor [n*r];
b[n_]:=n+Floor [n*s];
Table[a[n],{n,1,120}]  (* A184808 *)
Table[b[n],{n,1,120}]  (* A184809 *)
  • PARI
    main(size)={return(vector(size, n, n+floor(sqrt(2/3)*n)))} /* Anders Hellström, Jul 15 2015 */

Formula

a(n) = n + floor(r*n), where r = sqrt(2/3).

A190002 a(n) = n + [n*s/r] + [n*t/r]; r=1, s=(sinh(1))^2, t=(cosh(1))^2.

Original entry on oeis.org

4, 8, 14, 18, 22, 28, 32, 38, 42, 46, 52, 56, 60, 66, 70, 76, 80, 84, 90, 94, 100, 104, 108, 114, 118, 122, 128, 132, 138, 142, 146, 152, 156, 160, 166, 170, 176, 180, 184, 190, 194, 200, 204, 208, 214, 218, 222, 228, 232, 238, 242, 246, 252, 256, 260, 266, 270, 276, 280, 284, 290, 294, 300, 304, 308, 314, 318, 322, 328, 332, 338
Offset: 1

Views

Author

Clark Kimberling, May 03 2011

Keywords

Comments

This is one of three sequences that partition the positive integers. In general, suppose that r, s, t are positive real numbers for which the sets {i/r: i>=1}, {j/s: j>=1}, {k/t: k>=1} are pairwise disjoint. Let a(n) be the rank of n/r when all the numbers in the three sets are jointly ranked. Define b(n) and c(n) as the ranks of n/s and n/t. It is easy to prove that
a(n) = n + [n*s/r] + [n*t/r],
b(n) = n + [n*r/s] + [n*t/s],
c(n) = n + [n*r/t] + [n*s/t], where []=floor.
Taking r=1, s=(sinh(1))^2, t=(cosh(1))^2 gives
a=A190002, b=A190003, c=A190004.

Links

Crossrefs

Cf. A189999, A190003, A005408.

Programs

  • Magma
    [n + Floor(n*(Sinh(1))^2) + Floor(n*(Cosh(1))^2): n in [1..100]]; // G. C. Greubel, Jan 11 2018
  • Mathematica
    r=1; s=Sinh[1]^2; t=Cosh[1]^2;
a[n_] := n + Floor[n*s/r] + Floor[n*t/r];
b[n_] := n + Floor[n*r/s] + Floor[n*t/s];
c[n_] := n + Floor[n*r/t] + Floor[n*s/t];
Table[a[n], {n, 1, 120}]  (* A190002 *)
Table[b[n], {n, 1, 120}]  (* A190003 *)
Table[c[n], {n, 1, 120}]  (* A005408 *)
Table[a[n]/2, {n, 1, 120}](* A190004 *)
Table[b[n]/2, {n, 1, 120}](* A182760 *)
  • PARI
    for(n=1,100, print1(n + floor(n*(sinh(1))^2) + floor(n*(cosh(1))^2), ", ")) \\ G. C. Greubel, Jan 11 2018

Formula

A190002: a(n) = n + [n*(sinh(1))^2] + [n*(cosh(1))^2].
A190003: b(n) = n + [n*(csch(1))^2] + [n*(coth(1))^2].
A005408: c(n) = 2*n - 1.

A182774 Beatty sequence for 1+2^(-2/3).

Original entry on oeis.org

1, 3, 4, 6, 8, 9, 11, 13, 14, 16, 17, 19, 21, 22, 24, 26, 27, 29, 30, 32, 34, 35, 37, 39, 40, 42, 44, 45, 47, 48, 50, 52, 53, 55, 57, 58, 60, 61, 63, 65, 66, 68, 70, 71, 73, 74, 76, 78, 79, 81, 83, 84, 86, 88, 89, 91, 92, 94, 96, 97, 99, 101, 102, 104
Offset: 1

Views

Author

Clark Kimberling, Nov 30 2010

Keywords

Comments

Let u=2^(1/3). Jointly rank {ju} and {k/u} as in the first comment at A182760; a(n) is the position of n/u. A182774 is the complement of A182773.

Links

Crossrefs

Cf. A182760, A182773, A182774, A235362.

Programs

  • Magma
    [Floor(n*(1+2^(-2/3))): n in [1..80]]; // Vincenzo Librandi, Oct 25 2011
  • Mathematica
    Floor[Range[100]*(1 + 2^(-2/3))] (* Paolo Xausa, Jul 09 2024 *)

Formula

a(n) = floor(n*(1 + 2^(-2/3))).

A182777 Beatty sequence for 3-sqrt(3).

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 15, 16, 17, 19, 20, 21, 22, 24, 25, 26, 27, 29, 30, 31, 32, 34, 35, 36, 38, 39, 40, 41, 43, 44, 45, 46, 48, 49, 50, 51, 53, 54, 55, 57, 58, 59, 60, 62, 63, 64, 65, 67, 68, 69, 71, 72, 73, 74, 76, 77, 78, 79, 81, 82, 83, 84, 86
Offset: 1

Views

Author

Clark Kimberling, Nov 30 2010

Keywords

Comments

(1) 3 is the only number x for which the numbers r=x-sqrt(x) and s=x+sqrt(x) satisfy the Beatty equation
1/r + 1/s = 1.
(2) Let u=2-sqrt(3) and v=1. Jointly rank {j*u} and {k*v} as in the first comment at A182760; a(n) is the position of n*u.
(3) The complement of A182777 is A182778, which gives the positions of the natural numbers k in the joint ranking.

Links

Crossrefs

Cf. A182760, A182778.

Programs

  • Magma
    [Floor(n*(3-Sqrt(3))): n in [1..80]]; // Vincenzo Librandi, Oct 25 2011
  • Mathematica
    Table[Floor[(3-Sqrt[3]) n], {n, 68}]
  • PARI
    vector(80, n, floor(n*(3-sqrt(3)))) \\ G. C. Greubel, Nov 23 2018
  • Sage
    [floor(n*(3-sqrt(3))) for n in (1..80)] # G. C. Greubel, Nov 23 2018

Formula

a(n) = floor(n*(3-sqrt(3))).

Extensions

Typo in formula by Vincenzo Librandi, Oct 25 2011

A190003 a(n) = n + [n*r/s] + [n*t/s]; r=1, s=(sinh(1))^2, t=(cosh(1))^2.

Original entry on oeis.org

2, 6, 10, 12, 16, 20, 24, 26, 30, 34, 36, 40, 44, 48, 50, 54, 58, 62, 64, 68, 72, 74, 78, 82, 86, 88, 92, 96, 98, 102, 106, 110, 112, 116, 120, 124, 126, 130, 134, 136, 140, 144, 148, 150, 154, 158, 162, 164, 168, 172, 174, 178, 182, 186, 188, 192, 196, 198, 202, 206, 210, 212, 216, 220, 224, 226, 230, 234, 236, 240
Offset: 1

Views

Author

Clark Kimberling, May 03 2011

Keywords

Comments

See A190002.

Links

Crossrefs

Cf. A190002, A182760.

Programs

  • Magma
    [n + Floor(n/(Sinh(1))^2) + Floor(n/(Tanh(1))^2): n in [1..100]]; // G. C. Greubel, Jan 11 2018
  • Maple
    seq(n+floor(n/s)+floor(n*t/s), n=1..100); # Robert Israel, Jan 12 2018
  • Mathematica
    r=1; s=Sinh[1]^2; t=Cosh[1]^2;
a[n_] := n + Floor[n*s/r] + Floor[n*t/r];
b[n_] := n + Floor[n*r/s] + Floor[n*t/s];
c[n_] := n + Floor[n*r/t] + Floor[n*s/t];
Table[a[n], {n, 1, 120}]  (* A190002 *)
Table[b[n], {n, 1, 120}]  (* A190003 *)
Table[c[n], {n, 1, 120}]  (* A005408 *)
Table[a[n]/2, {n, 1, 120}](* A190004 *)
Table[b[n]/2, {n, 1, 120}](* A182760 *)
  • PARI
    for(n=1,100, print1(n + floor(n/(sinh(1))^2) + floor(n/(tanh(1))^2), ", ")) \\ G. C. Greubel, Jan 11 2018

Formula

A190002: a(n) = n + [n*(sinh(1))^2] + [n*(cosh(1))^2].
A190003: b(n) = n + [n*(csch(1))^2] + [n*(coth(1))^2].
A005408: c(n) = 2*n - 1.
Showing 1-10 of 19 results. Next

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