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.

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A294401 Solution of the complementary equation a(n) = a(n-1) + b(n-2) + 2n, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 9, 19, 32, 48, 67, 89, 115, 144, 176, 211, 249, 290, 334, 381, 431, 485, 542, 602, 665, 731, 800, 872, 947, 1025, 1106, 1190, 1277, 1368, 1462, 1559, 1659, 1762, 1868, 1977, 2089, 2204, 2322, 2443, 2567, 2694, 2824, 2957, 3094, 3234, 3377, 3523, 3672
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 A022940 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) + 4 = 9.
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 10, 11, 12, 13, 15, 16, ...)

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] + b[n - 2] + 2 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}]  (* A294401 *)
Table[b[n], {n, 0, 10}]

A294415 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + 1, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 11, 24, 47, 85, 148, 251, 419, 693, 1138, 1859, 3027, 4918, 7979, 12933, 20950, 33923, 54915, 88882, 143843, 232774, 376669, 609497, 986222, 1595777, 2582059, 4177898, 6760021, 10937985, 17698074, 28636129, 46334275, 74970478, 121304829, 196275385
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) + b(0) + 1 = 11
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14,...)

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] + b[n - 2] + 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}]  (* A294415 *)
Table[b[n], {n, 0, 10}]

A294416 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + n, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 12, 27, 54, 99, 174, 297, 498, 825, 1357, 2220, 3618, 5882, 9547, 15479, 25079, 40614, 65752, 106428, 172245, 278741, 451057, 729872, 1181007, 1910961, 3092053, 5003102, 8095246, 13098442, 21193785, 34292327, 55486215, 89778648, 145264972, 235043732
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) + b(0) + 2 = 12
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14,...)

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] + b[n - 2] + 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}]  (* A294416 *)
Table[b[n], {n, 0, 10}]

A294417 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - n, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 8, 17, 32, 57, 99, 168, 280, 462, 757, 1235, 2009, 3262, 5291, 8575, 13889, 22488, 36402, 58916, 95345, 154289, 249663, 403982, 653676, 1057690, 1711399, 2769123, 4480558, 7249719, 11730316, 18980075, 30710432, 49690549, 80401024, 130091617
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) + b(0) - 2 = 8
Complement: (b(n)) = (2, 4, 5, 6, 7, 9, 10, 11, 13, 14,...)

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] + b[n - 2] - 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}]  (* A294417 *)
Table[b[n], {n, 0, 10}]

A294418 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + 2*b(n-2), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 12, 28, 56, 103, 181, 309, 518, 858, 1411, 2309, 3763, 6118, 9930, 16100, 26085, 42243, 68389, 110696, 179152, 289918, 469143, 759137, 1228359, 1987579, 3216026, 5203696, 8419816, 13623609, 22043525, 35667237, 57710868, 93378214, 151089194, 244467523
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) + 2*b(0) = 12
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14,...)

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] + 2 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}]  (* A294418 *)
Table[b[n], {n, 0, 10}]

A294419 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + 2*b(n-1) + 2*b(n-2), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 16, 37, 75, 138, 243, 415, 696, 1153, 1895, 3098, 5047, 8203, 13314, 21587, 34975, 56640, 91697, 148423, 240210, 388727, 629035, 1017864, 1647005, 2664979, 4312098, 6977195, 11289415, 18266736, 29556281, 47823151, 77379570, 125202863, 202582581
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) + 2*b(1) + 2*b(0) = 16
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 17,...)

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] + 2 b[n - 1] + 2 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}]  (* A294419 *)
Table[b[n], {n, 0, 10}]

A294420 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + 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, 14, 31, 62, 113, 198, 337, 564, 933, 1532, 2503, 4078, 6628, 10756, 17437, 28249, 45745, 74056, 119866, 193990, 313927, 507991, 821995, 1330066, 2152144, 3482296, 5634529, 9116919, 14751546, 23868566, 38620216, 62488889, 101109215, 163598217, 264707548
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) + 2*b(1) + b(n-2) = 14
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 17,...)

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] + 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}]  (* A294420 *)
Table[b[n], {n, 0, 10}]

A294421 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + 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, 19, 36, 63, 108, 181, 302, 496, 812, 1323, 2151, 3491, 5660, 9170, 14852, 24044, 38919, 62987, 101931, 164944, 266902, 431874, 698805, 1130709, 1829545, 2960286, 4789864, 7750184, 12540083, 20290303, 32830425, 53120767, 85951232, 139072040
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) + 2*b(1) - b(0) = 10
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 11, 12, 13, ...)

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] + 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}]  (* A294421 *)
Table[b[n], {n, 0, 10}]

A294422 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) - b(n-2) + 1, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 7, 12, 21, 36, 59, 97, 158, 258, 418, 678, 1098, 1778, 2878, 4658, 7538, 12199, 19739, 31940, 51681, 83623, 135306, 218931, 354239, 573172, 927413, 1500587, 2428002, 3928591, 6356595, 10285189, 16641786, 26926977, 43568765, 70495744
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. 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) - b(0) + 1 = 7
Complement: (b(n)) = (2, 4, 5, 6, 8, 9, 11, 13, ...)

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] - b[n - 2] + 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}]  (* A294422 *)
Table[b[n], {n, 0, 10}]

A294423 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) - b(n-2) + n, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 8, 15, 28, 49, 85, 142, 236, 388, 635, 1035, 1684, 2733, 4432, 7181, 11630, 18829, 30478, 49327, 79826, 129175, 209024, 338223, 547273, 885522, 1432822, 2318372, 3751223, 6069625, 9820879, 15890536, 25711448, 41602018, 67313501, 108915555, 176229093
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. 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) - b(0)  + 2 = 8
Complement: (b(n)) = (2, 4, 5, 6, 7, 9, 11, 13, ...)

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] - b[n - 2] + 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}]  (* A294423 *)
Table[b[n], {n, 0, 10}]
Previous Showing 21-30 of 40 results. Next

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