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.

Previous Showing 21-30 of 63 results. Next

A329845 Beatty sequence for (3+sqrt(29))/5.

Original entry on oeis.org

1, 3, 5, 6, 8, 10, 11, 13, 15, 16, 18, 20, 21, 23, 25, 26, 28, 30, 31, 33, 35, 36, 38, 40, 41, 43, 45, 46, 48, 50, 51, 53, 55, 57, 58, 60, 62, 63, 65, 67, 68, 70, 72, 73, 75, 77, 78, 80, 82, 83, 85, 87, 88, 90, 92, 93, 95, 97, 98, 100, 102, 103, 105, 107
Offset: 1

Views

Author

Clark Kimberling, Jan 02 2020

Keywords

Comments

Let r = (3+sqrt(29))/5. Then (floor(n*r)) and (floor(n*r + 4r/5)) are a pair of Beatty sequences; i.e., every positive integer is in exactly one of the sequences. See the Guide to related sequences at A329825.

Links

Crossrefs

Cf. A329825, A329846 (complement).

Programs

  • Mathematica
    t = 4/5; r = Simplify[(2 - t + Sqrt[t^2 + 4])/2]; s = Simplify[r/(r - 1)];
Table[Floor[r*n], {n, 1, 200}]   (* A329845 *)
Table[Floor[s*n], {n, 1, 200}]   (* A329846 *)

Formula

a(n) = floor(n*r), where r = (3+sqrt(29))/5.

A329846 Beatty sequence for (7+sqrt(29))/5.

Original entry on oeis.org

2, 4, 7, 9, 12, 14, 17, 19, 22, 24, 27, 29, 32, 34, 37, 39, 42, 44, 47, 49, 52, 54, 56, 59, 61, 64, 66, 69, 71, 74, 76, 79, 81, 84, 86, 89, 91, 94, 96, 99, 101, 104, 106, 108, 111, 113, 116, 118, 121, 123, 126, 128, 131, 133, 136, 138, 141, 143, 146, 148
Offset: 1

Views

Author

Clark Kimberling, Jan 02 2020

Keywords

Comments

Let r = (3+sqrt(29))/5. Then (floor(n*r)) and (floor(n*r + 4r/5)) are a pair of Beatty sequences; i.e., every positive integer is in exactly one of the sequences. See the Guide to related sequences at A329825.

Links

Crossrefs

Cf. A329825, A329845 (complement).

Programs

  • Mathematica
    t = 4/5; r = Simplify[(2 - t + Sqrt[t^2 + 4])/2]; s = Simplify[r/(r - 1)];
Table[Floor[r*n], {n, 1, 200}]   (* A329845 *)
Table[Floor[s*n], {n, 1, 200}]   (* A329846 *)

Formula

a(n) = floor(n*s), where s = (7+sqrt(29))/5.

A329847 Beatty sequence for (3+sqrt(89))/8.

Original entry on oeis.org

1, 3, 4, 6, 7, 9, 10, 12, 13, 15, 17, 18, 20, 21, 23, 24, 26, 27, 29, 31, 32, 34, 35, 37, 38, 40, 41, 43, 45, 46, 48, 49, 51, 52, 54, 55, 57, 59, 60, 62, 63, 65, 66, 68, 69, 71, 73, 74, 76, 77, 79, 80, 82, 83, 85, 87, 88, 90, 91, 93, 94, 96, 97, 99, 101, 102
Offset: 1

Views

Author

Clark Kimberling, Jan 02 2020

Keywords

Comments

Let r = (3+sqrt(89))/8. Then (floor(n*r)) and (floor(n*r + 5r/4)) are a pair of Beatty sequences; i.e., every positive integer is in exactly one of the sequences. See the Guide to related sequences at A329825.

Links

Crossrefs

Cf. A329825, A329848 (complement).

Programs

  • Mathematica
    t = 5/4; r = Simplify[(2 - t + Sqrt[t^2 + 4])/2]; s = Simplify[r/(r - 1)];
Table[Floor[r*n], {n, 1, 200}]   (* A329847 *)
Table[Floor[s*n], {n, 1, 200}]   (* A329848 *)

Formula

a(n) = floor(n*r), where r = (3+sqrt(89))/8.

A329848 Beatty sequence for (13+sqrt(89))/8.

Original entry on oeis.org

2, 5, 8, 11, 14, 16, 19, 22, 25, 28, 30, 33, 36, 39, 42, 44, 47, 50, 53, 56, 58, 61, 64, 67, 70, 72, 75, 78, 81, 84, 86, 89, 92, 95, 98, 100, 103, 106, 109, 112, 114, 117, 120, 123, 126, 128, 131, 134, 137, 140, 143, 145, 148, 151, 154, 157, 159, 162, 165
Offset: 1

Views

Author

Clark Kimberling, Jan 02 2020

Keywords

Comments

Let r = (3+sqrt(89))/8. Then (floor(n*r)) and (floor(n*r + 5r/4)) are a pair of Beatty sequences; i.e., every positive integer is in exactly one of the sequences. See the Guide to related sequences at A329825.

Links

Crossrefs

Cf. A329825, A329847 (complement).

Programs

  • Mathematica
    t = 5/4; r = Simplify[(2 - t + Sqrt[t^2 + 4])/2]; s = Simplify[r/(r - 1)];
Table[Floor[r*n], {n, 1, 200}]   (* A329847 *)
Table[Floor[s*n], {n, 1, 200}]   (* A329848 *)

Formula

a(n) = floor(n*s), where s = (13+sqrt(89))/8.

A329923 Beatty sequence for (2+sqrt(34))/5.

Original entry on oeis.org

1, 3, 4, 6, 7, 9, 10, 12, 14, 15, 17, 18, 20, 21, 23, 25, 26, 28, 29, 31, 32, 34, 36, 37, 39, 40, 42, 43, 45, 46, 48, 50, 51, 53, 54, 56, 57, 59, 61, 62, 64, 65, 67, 68, 70, 72, 73, 75, 76, 78, 79, 81, 83, 84, 86, 87, 89, 90, 92, 93, 95, 97, 98, 100, 101
Offset: 1

Views

Author

Clark Kimberling, Jan 02 2020

Keywords

Comments

Let r = (2+sqrt(34))/5. Then (floor(n*r)) and (floor(n*r + 6r/5)) are a pair of Beatty sequences; i.e., every positive integer is in exactly one of the sequences. See the Guide to related sequences at A329825.

Links

Crossrefs

Cf. A329825, A329924 (complement).

Programs

  • Mathematica
    t = 6/5; r = Simplify[(2 - t + Sqrt[t^2 + 4])/2]; s = Simplify[r/(r - 1)];
Table[Floor[r*n], {n, 1, 200}]   (* A329923 *)
Table[Floor[s*n], {n, 1, 200}]   (* A329924 *)

Formula

a(n) = floor(n*r), where r = (2+sqrt(34))/5.

A329924 Beatty sequence for (8+sqrt(34))/5.

Original entry on oeis.org

2, 5, 8, 11, 13, 16, 19, 22, 24, 27, 30, 33, 35, 38, 41, 44, 47, 49, 52, 55, 58, 60, 63, 66, 69, 71, 74, 77, 80, 82, 85, 88, 91, 94, 96, 99, 102, 105, 107, 110, 113, 116, 118, 121, 124, 127, 130, 132, 135, 138, 141, 143, 146, 149, 152, 154, 157, 160, 163
Offset: 1

Views

Author

Clark Kimberling, Jan 02 2020

Keywords

Comments

Let r = (2+sqrt(34))/5. Then (floor(n*r)) and (floor(n*r + 6r/5)) are a pair of Beatty sequences; i.e., every positive integer is in exactly one of the sequences. See the Guide to related sequences at A329825.

Links

Crossrefs

Cf. A329825, A329923 (complement).

Programs

  • Mathematica
    t = 6/5; r = Simplify[(2 - t + Sqrt[t^2 + 4])/2]; s = Simplify[r/(r - 1)];
Table[Floor[r*n], {n, 1, 200}]   (* A329923 *)
Table[Floor[s*n], {n, 1, 200}]   (* A329924 *)

Formula

a(n) = floor(n*s), where s = (8+sqrt(34))/5.

A329925 Beatty sequence for (1+sqrt(41))/5.

Original entry on oeis.org

1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 25, 26, 28, 29, 31, 32, 34, 35, 37, 38, 39, 41, 42, 44, 45, 47, 48, 50, 51, 53, 54, 56, 57, 59, 60, 62, 63, 65, 66, 68, 69, 71, 72, 74, 75, 76, 78, 79, 81, 82, 84, 85, 87, 88, 90, 91, 93, 94, 96, 97
Offset: 1

Views

Author

Clark Kimberling, Jan 02 2020

Keywords

Comments

Let r = (1+sqrt(41))/5. Then (floor(n*r)) and (floor(n*r + 8r/5)) are a pair of Beatty sequences; i.e., every positive integer is in exactly one of the sequences. See the Guide to related sequences at A329825.

Links

Crossrefs

Cf. A329825, A329926 (complement).

Programs

  • Mathematica
    t = 8/5; r = Simplify[(2 - t + Sqrt[t^2 + 4])/2]; s = Simplify[r/(r - 1)];
Table[Floor[r*n], {n, 1, 200}]   (* A329925 *)
Table[Floor[s*n], {n, 1, 200}]   (* A329926 *)

Formula

a(n) = floor(n*r), where r = (1+sqrt(41))/5.

A329926 Beatty sequence for (9+sqrt(41))/5.

Original entry on oeis.org

3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 77, 80, 83, 86, 89, 92, 95, 98, 101, 104, 107, 110, 113, 117, 120, 123, 126, 129, 132, 135, 138, 141, 144, 147, 150, 154, 157, 160, 163, 166, 169, 172, 175, 178, 181
Offset: 1

Views

Author

Clark Kimberling, Jan 02 2020

Keywords

Comments

Let r = (1+sqrt(41))/5. Then (floor(n*r)) and (floor(n*r + 8r/5)) are a pair of Beatty sequences; i.e., every positive integer is in exactly one of the sequences. See the Guide to related sequences at A329825.

Links

Crossrefs

Cf. A329825, A329925 (complement).

Programs

  • Mathematica
    t = 8/5; r = Simplify[(2 - t + Sqrt[t^2 + 4])/2]; s = Simplify[r/(r - 1)];
Table[Floor[r*n], {n, 1, 200}]   (* A329925 *)
Table[Floor[s*n], {n, 1, 200}]   (* A329926 *)

Formula

a(n) = floor(n*s), where s = (9+sqrt(41))/5.

Extensions

Definition corrected by Georg Fischer, Jul 08 2021

A329938 Beatty sequence for sinh x, where csch x + sech x = 1 .

Original entry on oeis.org

1, 3, 5, 7, 9, 11, 13, 15, 16, 18, 20, 22, 24, 26, 28, 30, 32, 33, 35, 37, 39, 41, 43, 45, 47, 48, 50, 52, 54, 56, 58, 60, 62, 64, 65, 67, 69, 71, 73, 75, 77, 79, 80, 82, 84, 86, 88, 90, 92, 94, 96, 97, 99, 101, 103, 105, 107, 109, 111, 112, 114, 116, 118
Offset: 1

Views

Author

Clark Kimberling, Jan 02 2020

Keywords

Comments

Let x be the solution of csch x + sech x = 1. Then (floor(n*sinh x)) and (floor(n*cosh x)) are a pair of Beatty sequences; i.e., every positive integer is in exactly one of the sequences. See the Guide to related sequences at A329825.

Links

Crossrefs

Cf. A329825, A329831, A329937, A329939 (complement).

Programs

  • Mathematica
    Solve[1/Sinh[x] + 1/Cosh[x] == 1, x]
r = ArcSech[1/8 (4 - 4 Sqrt[2] - 9 Sqrt[5 + 4 Sqrt[2]] + (5 + 4 Sqrt[2])^(3/2))];
u = N[r, 250]
v = RealDigits[u][[1]];
Table[Floor[n*Sinh[r]], {n, 1, 150}]  (* A329938 *)
Table[Floor[n*Cosh[r]], {n, 1, 150}]  (* A329939 *)

Formula

a(n) = floor(n*sinh x), where x = 1.390148... is the constant in A329937; a(n) first differs from A329831(n) at n = 77.

A329939 Beatty sequence for cosh x, where csch x + sech x = 1 .

Original entry on oeis.org

2, 4, 6, 8, 10, 12, 14, 17, 19, 21, 23, 25, 27, 29, 31, 34, 36, 38, 40, 42, 44, 46, 49, 51, 53, 55, 57, 59, 61, 63, 66, 68, 70, 72, 74, 76, 78, 81, 83, 85, 87, 89, 91, 93, 95, 98, 100, 102, 104, 106, 108, 110, 113, 115, 117, 119, 121, 123, 125, 127, 130, 132
Offset: 1

Views

Author

Clark Kimberling, Jan 02 2020

Keywords

Comments

Let x be the solution of csch x + sech x = 1. Then (floor(n*sinh x)) and (floor(n*cosh x)) are a pair of Beatty sequences; i.e., every positive integer is in exactly one of the sequences. See the Guide to related sequences at A329825.

Links

Crossrefs

Cf. A329825, A329832, A329937, A329938 (complement).

Programs

  • Mathematica
    Solve[1/Sinh[x] + 1/Cosh[x] == 1, x]
r = ArcSech[1/8 (4 - 4 Sqrt[2] - 9 Sqrt[5 + 4 Sqrt[2]] + (5 + 4 Sqrt[2])^(3/2))];
u = N[r, 250]
v = RealDigits[u][[1]];
Table[Floor[n*Sinh[r]], {n, 1, 150}]  (* A329938 *)
Table[Floor[n*Cosh[r]], {n, 1, 150}]  (* A329939 *)

Formula

a(n) = floor(n*cosh x), where x = 1.390148... is the constant in A329937; a(n) first differs from A329832(n) at n = 68.
Previous Showing 21-30 of 63 results. Next

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.