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 40 results. Next

A294414 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 10, 22, 43, 78, 136, 231, 387, 641, 1053, 1721, 2803, 4555, 7391, 11981, 19409, 31429, 50879, 82352, 133278, 215679, 349008, 564740, 913803, 1478600, 2392462, 3871123, 6263648, 10134836, 16398551, 26533456, 42932078, 69465607, 112397760, 181863444
Offset: 0

Views

Author

Clark Kimberling, Oct 31 2017

Keywords

Comments

The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:
A294413: a(n) = a(n-1) + a(n-2) - b(n-1) + 6
A294414: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2)
A294415: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + 1
A294416: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + n
A294417: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - n
A294418: a(n) = a(n-1) + a(n-2) + b(n-1) + 2*b(n-2)
A294419: a(n) = a(n-1) + a(n-2) + 2*b(n-1) + 2*b(n-2)
A294420: a(n) = a(n-1) + a(n-2) + 2*b(n-1) + b(n-2)
A294421: a(n) = a(n-1) + a(n-2) + 2*b(n-1) - b(n-2)
A294422: a(n) = a(n-1) + a(n-2) + b(n-1) - b(n-2) + 1
A294423: a(n) = a(n-1) + a(n-2) + b(n-1) - b(n-2) + n
A294424: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - 1
A294425: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - 2
A294426: a(n) = a(n-1) + 2*a(n-2) + b(n-1) + b(n-2) - 3
Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio.

Examples

			a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
a(2)  = a(1) + a(0) + b(1) + b(0) = 6
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 11, 12, 13, ...)

Links

Crossrefs

Cf. A293076, A293765, A022940.

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + b[n - 2];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 40}]  (* A294414 *)
Table[b[n], {n, 0, 10}]

Extensions

Definition corrected by Georg Fischer, Sep 27 2020

A293358 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 8, 16, 30, 53, 92, 155, 258, 425, 696, 1135, 1846, 2998, 4862, 7879, 12761, 20661, 33444, 54128, 87596, 141749, 229371, 371147, 600546, 971722, 1572299, 2544053, 4116385, 6660472, 10776892, 17437400, 28214329, 45651767, 73866135, 119517942
Offset: 0

Views

Author

Clark Kimberling, Oct 29 2017

Keywords

Comments

The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:
A293358: a(n) = a(n-1) + a(n-2) + b(n-1)
A293406: a(n) = a(n-1) + a(n-2) + b(n-1) + 1
A293765: a(n) = a(n-1) + a(n-2) + b(n-1) + 2
A293766: a(n) = a(n-1) + a(n-2) + b(n-1) + 3
A293767: a(n) = a(n-1) + a(n-2) + b(n-1) - 1
A294365: a(n) = a(n-1) + a(n-2) + b(n-1) + n
A294366: a(n) = a(n-1) + a(n-2) + b(n-1) + 2n
A294367: a(n) = a(n-1) + a(n-2) + b(n-1) + n - 1
A294368: a(n) = a(n-1) + a(n-2) + b(n-1) + n + 1
Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio.

Examples

			a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
a(2) = a(1) + a(0) + b(1) = 8;
Complement: (b(n)) = (2, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, ...)

Links

Crossrefs

Cf. A001622 (golden ratio), A293076.

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 40}]  (* A293358 *)
Table[b[n], {n, 0, 10}]

A022940 a(n) = a(n-1) + b(n-2) for n >= 3, a( ) increasing, given a(1) = 1, a(2) = 3; where b( ) is complement of a( ).

Original entry on oeis.org

1, 3, 5, 9, 15, 22, 30, 40, 51, 63, 76, 90, 106, 123, 141, 160, 180, 201, 224, 248, 273, 299, 326, 354, 383, 414, 446, 479, 513, 548, 584, 621, 659, 698, 739, 781, 824, 868, 913, 959, 1006, 1054, 1103, 1153, 1205, 1258, 1312, 1367, 1423, 1480
Offset: 1

Views

Author

Clark Kimberling

Keywords

Comments

From Clark Kimberling, Oct 30 2017: (Start)
The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:
here: a(n) = a(n-1) + b(n-2) [with a different offset]
A294397: a(n) = a(n-1) + b(n-2) + 1;
A294398: a(n) = a(n-1) + b(n-2) + 2;
A294399: a(n) = a(n-1) + b(n-2) + 3;
A294400: a(n) = a(n-1) + b(n-2) + n;
A294401: a(n) = a(n-1) + b(n-2) + 2*n.
(End)

Examples

			a(1) = 1, a(2) = 3, b(1) = 2, b(2) = 4, so that a(3) = a(2) + a(1) + b(2) = 5.
Complement: {b(n)} = {2, 4, 6, 7, 8, 10, 11, 12, 13, 14, 16, ...}

Links

Crossrefs

Cf. A005228 and references therein.
Cf. A293076, A293765, A294381.

Programs

  • Mathematica
    Fold[Append[#1, #1[[-1]] + Complement[Range[Max@#1 + 1], #1][[#2]]] &, {1, 3}, Range[50]] (* Ivan Neretin, Apr 04 2016 *)

A294381 Solution of the complementary equation a(n) = a(n-1)*b(n-2), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 6, 24, 120, 840, 6720, 60480, 604800, 6652800, 79833600, 1037836800, 14529715200, 217945728000, 3487131648000, 59281238016000, 1067062284288000, 20274183401472000, 405483668029440000, 8515157028618240000, 187333454629601280000, 4308669456480829440000, 107716736412020736000000
Offset: 0

Views

Author

Clark Kimberling, Oct 29 2017

Keywords

Comments

The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:
A294381: a(n) = a(n-1)*b(n-2)
A294382: a(n) = a(n-1)*b(n-2) - 1
A294383: a(n) = a(n-1)*b(n-2) + 1
A294384: a(n) = a(n-1)*b(n-2) - n
A294385: a(n) = a(n-1)*b(n-2) + n

Examples

			a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that a(2) = a(1)*b(0) = 6.
Complement: (b(n)) = (2, 4, 5, 7, 8, 9, 10, 12, 13, 14, 15, 16, ...).

Links

Crossrefs

Cf. A293076, A293765.

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
a[n_] := a[n] = a[n - 1]*b[n - 2];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 40}]  (* A294381 *)
Table[b[n], {n, 0, 10}]

Extensions

More terms from Jack W Grahl, Apr 26 2018

A294476 Solution of the complementary equation a(n) = a(n-2) + b(n-1) + 1, where a(0) = 1, a(1) = 3, b(0) = 2.

Original entry on oeis.org

1, 3, 6, 9, 14, 18, 25, 30, 38, 44, 54, 61, 72, 81, 93, 103, 116, 127, 141, 154, 169, 183, 199, 215, 232, 249, 267, 285, 304, 323, 344, 364, 386, 407, 430, 453, 477, 501, 526, 551, 577, 603, 630, 657, 686, 714, 744, 773, 804, 834, 867, 898, 932, 964, 999
Offset: 0

Views

Author

Clark Kimberling, Nov 01 2017

Keywords

Comments

The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2:
A294476: a(n) = a(n-2) + b(n-1) + 1
A294477: a(n) = a(n-2) + b(n-1) + 2
A294478: a(n) = a(n-2) + b(n-1) + 3
A294479: a(n) = a(n-2) + b(n-1) + n
A294480: a(n) = a(n-2) + b(n-1) + 2n
A294481: a(n) = a(n-2) + b(n-1) + n - 1
A294482: a(n) = a(n-2) + b(n-1) + n + 1
For a(n-2) + b(n-1), with offset 1 instead of 0, see A022942.

Examples

			a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
a(2)  = a(0) + b(1) + 1 = 6
Complement: (b(n)) = (2, 4, 5, 7, 8, 10, 11, 12, 13, 15,...)

Links

Crossrefs

Cf. A293076, A293765, A293358, A294414.

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 3; b[0] = 2;
a[n_] := a[n] = a[n - 2] + b[n - 1] + 1;
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 40}]  (* A294476 *)
Table[b[n], {n, 0, 10}]

A294413 Solution of the complementary equation a(n) = a(n-1) + a(n-2) - b(n-1) + 6, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 6, 10, 15, 23, 35, 53, 82, 128, 202, 320, 511, 819, 1317, 2122, 3424, 5530, 8936, 14447, 23363, 37789, 61130, 98896, 160002, 258873, 418849, 677695, 1096516, 1774181, 2870666, 4644815, 7515448, 12160229, 19675642, 31835835
Offset: 0

Views

Author

Clark Kimberling, Oct 31 2017

Keywords

Comments

The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A294414 for a guide to related sequences.
Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio.

Examples

			a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
a(2)  = a(1) + a(0) - b(1) + 6
Complement: (b(n)) = (2, 4, 6, 7, 9, 11, 12, 13, 14, 16, ...)

Links

Crossrefs

Cf. A293076, A293765, A294414.

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
a[n_] := a[n] = a[n - 1] + a[n - 2] - b[n - 1] + 6;
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 40}]  (* A294413 *)
Table[b[n], {n, 0, 10}]

A293766 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + 3, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 11, 22, 42, 74, 127, 213, 353, 581, 950, 1548, 2516, 4083, 6619, 10723, 17364, 28110, 45498, 73634, 119159, 192821, 312009, 504860, 816900, 1321792, 2138725, 3460551, 5599311, 9059898, 14659246, 23719182, 38378467, 62097689, 100476197, 162573928
Offset: 0

Views

Author

Clark Kimberling, Oct 29 2017

Keywords

Comments

The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:
Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio. See A293358 for a guide to related sequences.

Examples

			a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
a(2) = a(1) + a(0) + b(1) + 3 = 11;
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, ...)

Links

Crossrefs

Cf. A001622 (golden ratio), A293765.

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + 3;
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 40}]  (* A293766 *)
Table[b[n], {n, 0, 10}]

A293767 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) - 1, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 7, 14, 26, 47, 81, 137, 228, 376, 616, 1006, 1637, 2659, 4313, 6990, 11322, 18332, 29675, 48029, 77727, 125780, 203533, 329340, 532901, 862270, 1395201, 2257502, 3652735, 5910270, 9563039, 15473344, 25036419, 40509800, 65546257, 106056096
Offset: 0

Views

Author

Clark Kimberling, Oct 29 2017

Keywords

Comments

The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:
Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio. See A293358 for a guide to related sequences.

Examples

			a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
a(2) = a(1) + a(0) + b(1) - 1 = 7;
Complement: (b(n)) = (2, 4, 5, 6, 8, 9, 10, 11, 12, 13, 15, 16, 17, ...)

Links

Crossrefs

Cf. A001622 (golden ratio), A293765.

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] - 1;
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 40}]  (* A293767 *)
Table[b[n], {n, 0, 10}]

A294365 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + n, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 10, 21, 41, 74, 129, 219, 367, 607, 997, 1629, 2653, 4311, 6995, 11339, 18369, 29745, 48154, 77941, 126139, 204126, 330313, 534489, 864854, 1399397, 2264307, 3663762, 5928129, 9591953, 15520146, 25112165, 40632379, 65744614, 106377065, 172121753
Offset: 0

Views

Author

Clark Kimberling, Oct 29 2017

Keywords

Comments

The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:
Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio. See A293358 for a guide to related sequences.

Examples

			a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
a(2)  = a(1) + a(0) + b(1) + 2 = 10;
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, ...)

Links

Crossrefs

Cf. A001622 (golden ratio), A293765.

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + n;
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 40}]  (* A294365 *)
Table[b[n], {n, 0, 10}]

A294366 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + 2n, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 12, 26, 52, 95, 167, 285, 478, 792, 1303, 2131, 3473, 5646, 9164, 14858, 24073, 38985, 63115, 102160, 165338, 267564, 432971, 700608, 1133655, 1834342, 2968079, 4802506, 7770673, 12573270, 20344037, 32917404, 53261541, 86179048, 139440695, 225619852
Offset: 0

Views

Author

Clark Kimberling, Oct 29 2017

Keywords

Comments

The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:
Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio. See A293358 for a guide to related sequences.

Examples

			a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
a(2)  = a(1) + a(0) + b(1) + 4 = 12;
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, ...)

Links

Crossrefs

Cf. A001622 (golden ratio), A293765.

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + 2n;
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 40}]  (* A294366 *)
Table[b[n], {n, 0, 10}]
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