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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A295061 Solution of the complementary equation a(n) = 4*a(n-2) + b(n-1), where a(0) = 1, a(1) = 3, b(0) = 2, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 3, 8, 17, 38, 75, 161, 310, 655, 1252, 2633, 5022, 10547, 20104, 42206, 80435, 168844, 321761, 675398, 1287067, 2701616, 5148293, 10806490, 20593199, 43225988, 82372825, 172903982
Offset: 0

Views

Author

Clark Kimberling, Nov 18 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295053 for a guide to related sequences.
The sequence a(n+1)/a(n) appears to have two convergent subsequences, with limits 1.09... and 2.09... .

Examples

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

Crossrefs

Cf. A295053.

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
    a[0] = 1; a[1] = 3; b[0] = 2;
    a[n_] := a[n] = 4 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, 18}]  (* A295061 *)
    Table[b[n], {n, 0, 10}]

A295062 Solution of the complementary equation a(n) = 4*a(n-2) + b(n-2), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 3, 6, 16, 29, 71, 124, 293, 506, 1183, 2036, 4745, 8158, 18995, 32649, 75998, 130615, 304012, 522481, 1216070, 2089947, 4864304, 8359813, 19457242, 33439279, 77828996, 133757146, 311316015
Offset: 0

Views

Author

Clark Kimberling, Nov 18 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295053 for a guide to related sequences.
The sequence a(n+1)/a(n) appears to have two convergent subsequences, with limits 1.71... and 2.32... .

Examples

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

Crossrefs

Cf. A295053.

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] = 4 a[n - 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, 18}]  (* A295062 *)
    Table[b[n], {n, 0, 10}]

A295063 Solution of the complementary equation a(n) = 4*a(n-2) + b(n-1) + b(n-2), where a(0) = 1, a(1) = 3, b(0) = 2, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 3, 10, 21, 51, 97, 219, 405, 896, 1643, 3609, 6599, 14465, 26427, 57893, 105743, 231609, 423011, 926478, 1692089, 3705959, 6768405, 14823887, 27073673, 59295603, 108294749, 237182471
Offset: 0

Views

Author

Clark Kimberling, Nov 19 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295053 for a guide to related sequences.
The sequence a(n+1)/a(n) appears to have two convergent subsequences, with limits 2.19... and 1.82... .

Examples

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

Crossrefs

Cf. A295053.

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] = 4 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, 18}]  (* A295063 *)
    Table[b[n], {n, 0, 10}]

A295065 Solution of the complementary equation a(n) = 8*a(n-3) + b(n-2), where a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 3, 5, 12, 30, 47, 104, 249, 386, 843, 2005, 3102, 6759, 16056, 24833, 54090, 128467, 198684, 432741, 1027758, 1589495, 3461952, 8222089, 12715986, 27695643, 65776740, 101727917
Offset: 0

Views

Author

Clark Kimberling, Nov 19 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295053 for a guide to related sequences.
The sequence a(n+1)/a(n) appears to have three convergent subsequences, with limits 1.54..., 2.17..., 2.37...

Examples

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

Crossrefs

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
    a[0] = 1; a[1] = 3; a[2] = 5; b[0] = 2;
    a[n_] := a[n] = 8 a[n - 3] + 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, 18}]  (* A295065 *)
    Table[b[n], {n, 0, 10}]

A295070 Solution of the complementary equation a(n) = a(n-2) + b(n-1) + b(n-2), where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 2, 8, 11, 19, 24, 35, 43, 57, 68, 84, 97, 115, 130, 150, 168, 191, 211, 236, 259, 287, 312, 342, 369, 401, 430, 464, 495, 531, 565, 604, 640, 681, 719, 762, 802, 848, 891, 939, 984, 1034, 1081, 1133, 1182, 1236, 1287, 1343, 1396, 1454, 1510, 1571, 1629
Offset: 0

Views

Author

Clark Kimberling, Nov 19 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295053 for a guide to related sequences. Conjecture: a(n)/a(n-1) -> 1.

Examples

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

Crossrefs

Cf. A295053.

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
    a[0] = 1; a[1] = 2; b[0] = 3;
    a[n_] := a[n] = 2 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, 18}]  (* A295070 *)
    Table[b[n], {n, 0, 10}]

A295133 Solution of the complementary equation a(n) = 3*a(n-1) + b(n-1), where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 2, 10, 35, 111, 340, 1028, 3093, 9290, 27882, 83659, 250991, 752988, 2258980, 6776957, 20330889, 60992686, 182978078, 548934255
Offset: 0

Views

Author

Clark Kimberling, Nov 19 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295053 for a guide to related sequences.

Examples

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

Crossrefs

Cf. A295053.

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
    a[0] = 1; a[1] = 2; b[0] = 3;
    a[n_] := a[n] = 3 a[n - 1] + 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, 18}]  (* A295133 *)
    Table[b[n], {n, 0, 10}]

A295134 Solution of the complementary equation a(n) = 3*a(n-1) + b(n-1) - 1, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 2, 9, 31, 98, 300, 907, 2730, 8200, 24611, 73845, 221548, 664658, 1993989, 5981983, 17945966, 53837916, 161513767
Offset: 0

Views

Author

Clark Kimberling, Nov 19 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295053 for a guide to related sequences.

Examples

			a(0) = 1, a(1) = 2, b(0) = 3
a(2) =3*a(1) + b(1) - 1 = 9
Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, ... )
		

Crossrefs

Cf. A295053.

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
    a[0] = 1; a[1] = 2; b[0] = 3;
    a[n_] := a[n] = 3 a[n - 1] + 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, 18}]  (* A295134 *)
    Table[b[n], {n, 0, 10}]

A295135 Solution of the complementary equation a(n) = 3*a(n-1) + b(n-1) - 2, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 2, 8, 27, 85, 260, 787, 2369, 7116, 21358, 64085, 192267, 576814, 1730456, 5191383, 15574165, 46722512, 140167554
Offset: 0

Views

Author

Clark Kimberling, Nov 19 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295053 for a guide to related sequences.

Examples

			a(0) = 1, a(1) = 2, b(0) = 3
a(2) =3*a(1) + b(1) - 2 = 8
Complement: (b(n)) = (3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, ... )
		

Crossrefs

Cf. A295053.

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
    a[0] = 1; a[1] = 2; b[0] = 3;
    a[n_] := a[n] = 3 a[n - 1] + b[n - 1] -2;
    b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
    Table[a[n], {n, 0, 18}]  (* A295135 *)
    Table[b[n], {n, 0, 10}]

A295136 Solution of the complementary equation a(n) = 3*a(n-1) + b(n-1) - 3, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 2, 7, 23, 72, 221, 669, 2014, 6050, 18159, 54487, 163472, 490428, 1471297, 4413905, 13241730, 39725206, 119175635, 357526923
Offset: 0

Views

Author

Clark Kimberling, Nov 19 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295053 for a guide to related sequences.

Examples

			a(0) = 1, a(1) = 2, b(0) = 3
a(2) =3*a(1) + b(1) - 3 = 7
Complement: (b(n)) = (3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 14, ... )
		

Crossrefs

Cf. A295053.

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
    a[0] = 1; a[1] = 2; b[0] = 3;
    a[n_] := a[n] = 3 a[n - 1] + 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, 18}]  (* A295136 *)
    Table[b[n], {n, 0, 10}]

A295137 Solution of the complementary equation a(n) = 3*a(n-1) + b(n-1) - n, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 2, 8, 26, 80, 242, 729, 2190, 6573, 19722, 59169, 177510, 532533, 1597602, 4792809, 14378430, 43135293, 129405882
Offset: 0

Views

Author

Clark Kimberling, Nov 19 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295053 for a guide to related sequences.

Examples

			a(0) = 1, a(1) = 2, b(0) = 3
a(2) =3*a(1) + b(1) - n = 8
Complement: (b(n)) = (3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, ... )
		

Crossrefs

Cf. A295053.

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
    a[0] = 1; a[1] = 2; b[0] = 3;
    a[n_] := a[n] = 3 a[n - 1] + 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, 18}]  (* A295137 *)
    Table[b[n], {n, 0, 10}]
Previous Showing 21-30 of 33 results. Next