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 31-40 of 63 results. Next

A329961 Beatty sequence for 2 + sin x, where x = least positive solution of 1/(2 + sin x) + 1/(2 + cos x) = 1.

Original entry on oeis.org

2, 5, 8, 11, 14, 17, 20, 23, 25, 28, 31, 34, 37, 40, 43, 46, 49, 51, 54, 57, 60, 63, 66, 69, 72, 74, 77, 80, 83, 86, 89, 92, 95, 98, 100, 103, 106, 109, 112, 115, 118, 121, 123, 126, 129, 132, 135, 138, 141, 144
Offset: 1

Views

Author

Clark Kimberling, Jan 02 2020

Keywords

Comments

Let x be the least positive solution of 1/(2 + sin x) + 1/(2 + cos x) = 1. Then (floor(n*(2 + sin x))) and (floor(n*(2 + cos 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, A329960, A329962 (complement).

Programs

  • Mathematica
    Solve[1/(2 + Sin[x]) + 1/(2 + Cos[x]) == 1, x]
u = ArcCos[-(1/2) + 1/Sqrt[2] - 1/2 Sqrt[-1 + 2 Sqrt[2]]]
u1 = N[u, 150]
RealDigits[u1, 10][[1]]  (* A329960 *)
Table[Floor[n*(2 + Sin[u])], {n, 1, 50}]  (* A329961 *)
Table[Floor[n*(2 + Cos[u])], {n, 1, 50}]  (* A329962 *)
Plot[1/(2 + Sin[x]) + 1/(2 + Cos[x]) - 1, {x, -1, 3}]

Formula

a(n) = floor(n*(2 + sin x)), where x = 2.058943... is the constant in A329960.

A329962 Beatty sequence for 2 + cos x, where x = least positive solution of 1/(2 + sin x) + 1/(2 + cos x) = 1.

Original entry on oeis.org

1, 3, 4, 6, 7, 9, 10, 12, 13, 15, 16, 18, 19, 21, 22, 24, 26, 27, 29, 30, 32, 33, 35, 36, 38, 39, 41, 42, 44, 45, 47, 48, 50, 52, 53, 55, 56, 58, 59, 61, 62, 64, 65, 67, 68, 70, 71, 73, 75, 76, 78, 79, 81, 82, 84, 85, 87, 88, 90, 91, 93, 94, 96, 97, 99, 101
Offset: 1

Views

Author

Clark Kimberling, Jan 02 2020

Keywords

Comments

Let x be the least positive solution of 1/(2 + sin x) + 1/(2 + cos x) = 1. Then (floor(n*(2 + sin x))) and (floor(n*(2 + cos 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, A329960, A329961 (complement).

Programs

  • Mathematica
    Solve[1/(2 + Sin[x]) + 1/(2 + Cos[x]) == 1, x]
u = ArcCos[-(1/2) + 1/Sqrt[2] - 1/2 Sqrt[-1 + 2 Sqrt[2]]]
u1 = N[u, 150]
RealDigits[u1, 10][[1]]  (* A329960 *)
Table[Floor[n*(2 + Sin[u])], {n, 1, 50}]  (* A329961 *)
Table[Floor[n*(2 + Cos[u])], {n, 1, 50}]  (* A329962 *)
Plot[1/(2 + Sin[x]) + 1/(2 + Cos[x]) - 1, {x, -1, 3}]

Formula

a(n) = floor(n*(2 + cos x)), where x = 2.058943... is the constant in A329960.

A329977 Beatty sequence for the number x satisfying 1/x + 1/(log x) = 1.

Original entry on oeis.org

3, 7, 11, 15, 19, 23, 27, 30, 34, 38, 42, 46, 50, 54, 57, 61, 65, 69, 73, 77, 81, 84, 88, 92, 96, 100, 104, 108, 111, 115, 119, 123, 127, 131, 135, 138, 142, 146, 150, 154, 158, 162, 165, 169, 173, 177, 181, 185, 189, 192
Offset: 1

Views

Author

Clark Kimberling, Jan 02 2020

Keywords

Comments

Let x be the real solution of 1/x + 1/(log x) = 1. Then (floor(n x)) and (floor(n*(log(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, A236229, A329978 (complement).

Programs

  • Mathematica
    r = x /. FindRoot[1/x + 1/Log[x] == 1, {x, 3, 4}, WorkingPrecision -> 210];
RealDigits[r][[1]]; (* A236229 *)
Table[Floor[n*r], {n, 1, 50}];       (* A329977 *)
Table[Floor[n*Log[r]], {n, 1, 50}];  (* A329978 *)

Formula

a(n) = floor(n x), where x = 3.8573348... is the constant in A236229.

A329978 Beatty sequence for log x, where 1/x + 1/(log x) = 1.

Original entry on oeis.org

1, 2, 4, 5, 6, 8, 9, 10, 12, 13, 14, 16, 17, 18, 20, 21, 22, 24, 25, 26, 28, 29, 31, 32, 33, 35, 36, 37, 39, 40, 41, 43, 44, 45, 47, 48, 49, 51, 52, 53, 55, 56, 58, 59, 60, 62, 63, 64, 66, 67
Offset: 1

Views

Author

Clark Kimberling, Jan 02 2020

Keywords

Comments

Let x be the real solution of 1/x + 1/(log x) = 1. Then (floor(n x)) and (floor(n*(log(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. A236229, A329825, A329977 (complement).

Programs

  • Mathematica
    r = x /. FindRoot[1/x + 1/Log[x] == 1, {x, 3, 4}, WorkingPrecision -> 210];
RealDigits[r][[1]]; (* A236229 *)
Table[Floor[n*r], {n, 1, 50}];       (* A329977 *)
Table[Floor[n*Log[r]], {n, 1, 50}];  (* A329978 *)

Formula

a(n) = floor(n x), where x = 3.8573348... is the constant in A236229.

A329987 Beatty sequence for the number x satisfying 1/x + 1/2^x = 1.

Original entry on oeis.org

1, 3, 4, 6, 7, 9, 10, 12, 13, 15, 16, 18, 19, 21, 22, 24, 26, 27, 29, 30, 32, 33, 35, 36, 38, 39, 41, 42, 44, 45, 47, 48, 50, 52, 53, 55, 56, 58, 59, 61, 62, 64, 65, 67, 68, 70, 71, 73, 74, 76, 78, 79, 81, 82, 84, 85, 87, 88, 90, 91, 93, 94, 96, 97, 99, 100
Offset: 1

Views

Author

Clark Kimberling, Jan 02 2020

Keywords

Comments

Let x be the solution of 1/x + 1/2^x = 1. Then (floor(n x)) and (floor(n 2^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, A329986, A329988 (complement).

Programs

  • Mathematica
    r = x /. FindRoot[1/x + 1/2^x == 1, {x, 1, 10}, WorkingPrecision -> 210]
RealDigits[r][[1]] (* A329986 *)
Table[Floor[n*r], {n, 1, 250}]  (* A329987 *)
Table[Floor[n*2^r], {n, 1, 250}]  (* A329988 *)

Formula

a(n) = floor(n x), where x = 1.52980838275... is the constant in A329986.

A329988 Beatty sequence for 2^x, where 1/x + 1/2^x = 1.

Original entry on oeis.org

2, 5, 8, 11, 14, 17, 20, 23, 25, 28, 31, 34, 37, 40, 43, 46, 49, 51, 54, 57, 60, 63, 66, 69, 72, 75, 77, 80, 83, 86, 89, 92, 95, 98, 101, 103, 106, 109, 112, 115, 118, 121, 124, 127, 129, 132, 135, 138, 141, 144, 147, 150, 153, 155, 158, 161, 164, 167, 170
Offset: 1

Views

Author

Clark Kimberling, Jan 02 2020

Keywords

Comments

Let x be the solution of 1/x + 1/2^x = 1. Then (floor(n x)) and (floor(n 2^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, A329986, A329987 (complement).

Programs

  • Mathematica
    r = x /. FindRoot[1/x + 1/2^x == 1, {x, 1, 10}, WorkingPrecision -> 120]
RealDigits[r][[1]] (* A329986 *)
Table[Floor[n*r], {n, 1, 250}]    (* A329987 *)
Table[Floor[n*2^r], {n, 1, 250}]  (* A329988 *)

Formula

a(n) = floor(n 2^x), where x = 1.52980838275... is the constant in A329986.

A329990 Beatty sequence for the number x satisfying 1/x + 1/3^x = 1.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15, 17, 18, 19, 20, 22, 23, 24, 26, 27, 28, 30, 31, 32, 34, 35, 36, 38, 39, 40, 41, 43, 44, 45, 47, 48, 49, 51, 52, 53, 55, 56, 57, 58, 60, 61, 62, 64, 65, 66, 68, 69, 70, 72, 73, 74, 76, 77, 78, 79, 81, 82, 83, 85, 86
Offset: 1

Views

Author

Clark Kimberling, Jan 02 2020

Keywords

Comments

Let x be the solution of 1/x + 1/3^x = 1. Then (floor(n x)) and (floor(n 3^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, A329989, A329991 (complement).

Programs

  • Mathematica
    r = x /. FindRoot[1/x + 1/3^x == 1, {x, 1, 10}, WorkingPrecision -> 120]
RealDigits[r][[1]] (* A329989 *)
Table[Floor[n*r], {n, 1, 250}]  (* A329990 *)
Table[Floor[n*2^r], {n, 1, 250}]  (* A329991 *)

Formula

a(n) = floor(n x), where x = 1.31056994... is the constant in A329989.

A329991 Beatty sequence for 3^x, where 1/x + 1/3^x = 1.

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, 57, 59, 62, 64, 66, 69, 71, 74, 76, 79, 81, 84, 86, 89, 91, 94, 96, 99, 101, 104, 106, 109, 111, 114, 116, 119, 121, 124, 126, 128, 131, 133, 136, 138, 141, 143, 146, 148
Offset: 1

Views

Author

Clark Kimberling, Jan 02 2020

Keywords

Comments

Let x be the solution of 1/x + 1/3^x = 1. Then (floor(n x)) and (floor(n 3^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, A329989, A329990 (complement).

Programs

  • Mathematica
    r = x /. FindRoot[1/x + 1/3^x == 1, {x, 1, 10}, WorkingPrecision -> 120]
RealDigits[r][[1]] (* A329989 *)
Table[Floor[n*r], {n, 1, 250}]    (* A329990 *)
Table[Floor[n*2^r], {n, 1, 250}]  (* A329991 *)

Formula

a(n) = floor(n x), where x = 1.31056994... is the constant in A329989.

A329993 Beatty sequence for x^2, where 1/x^2 + 1/2^x = 1.

Original entry on oeis.org

1, 3, 5, 6, 8, 10, 11, 13, 15, 16, 18, 20, 21, 23, 25, 26, 28, 30, 32, 33, 35, 37, 38, 40, 42, 43, 45, 47, 48, 50, 52, 53, 55, 57, 58, 60, 62, 64, 65, 67, 69, 70, 72, 74, 75, 77, 79, 80, 82, 84, 85, 87, 89, 91, 92, 94, 96, 97, 99, 101, 102, 104, 106, 107
Offset: 1

Views

Author

Clark Kimberling, Jan 02 2020

Keywords

Comments

Let x be the solution of 1/x^2 + 1/2^x = 1. Then (floor(n x^2)) and (floor(n 2^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, A329992, A329994 (complement).

Programs

  • Mathematica
    r = x /. FindRoot[1/x^2 + 1/2^x == 1, {x, 1, 10}, WorkingPrecision -> 120]
RealDigits[r][[1]] (* A329992 *)
Table[Floor[n*r^2], {n, 1, 250}]  (* A329993 *)
Table[Floor[n*2^r], {n, 1, 250}]  (* A329994 *)

Formula

a(n) = floor(n*x^2), where x = 1.29819... is the constant in A329992; a(n) first differs from A064994(n) at n=89.

A329994 Beatty sequence for 2^x, where 1/x^2 + 1/2^x = 1.

Original entry on oeis.org

2, 4, 7, 9, 12, 14, 17, 19, 22, 24, 27, 29, 31, 34, 36, 39, 41, 44, 46, 49, 51, 54, 56, 59, 61, 63, 66, 68, 71, 73, 76, 78, 81, 83, 86, 88, 90, 93, 95, 98, 100, 103, 105, 108, 110, 113, 115, 118, 120, 122, 125, 127, 130, 132, 135, 137, 140, 142, 145, 147
Offset: 1

Views

Author

Clark Kimberling, Jan 02 2020

Keywords

Comments

Let x be the solution of 1/x^2 + 1/2^x = 1. Then (floor(n x^2)) and (floor(n 2^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, A329992, A329993 (complement).

Programs

  • Mathematica
    r = x /. FindRoot[1/x^2 + 1/2^x == 1, {x, 1, 10}, WorkingPrecision -> 120]
RealDigits[r][[1]] (* A329992 *)
Table[Floor[n*r^2], {n, 1, 250}]  (* A329993 *)
Table[Floor[n*2^r], {n, 1, 250}]  (* A329994 *)

Formula

a(n) = floor(n*2^x), where x = 1.298192... is the constant in A329992.

Extensions

Formula corrected. - R. J. Mathar, Jan 24 2020
Previous Showing 31-40 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.