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 1-4 of 4 results.

A295053 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(0) + b(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, 24, 52, 101, 186, 329, 568, 962, 1608, 2662, 4377, 7162, 11679, 18999, 30855, 50051, 81124, 131415, 212802, 344505, 557621, 902467, 1460457, 2363322, 3824207, 6187988, 10012686, 16201198, 26214442, 42416233, 68631304, 111048203
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. Guide to related sequencres:
A295053: a(n) = a(n-1) + a(n-2) + b(0) + b(1) + ... + b(n-1), a(0) = 1, a(1) = 2, b(0) = 3
A295054: a(n) = a(n-1) + a(n-2) + b(1) + b(2) + ... + b(n-1), a(0) = 1, a(1) = 2, b(0) = 3
A295055: a(n) = a(n-2) + b(1) + b(2) + ... + b(n-1), a(0) = 1, a(1) = 2, b(0) = 3
A295056: a(n) = 2*a(n-1) + b(n-1), a(0) = 1, a(1) = 4, b(0) = 2, b(1) = 3
A295057: a(n) = 2*a(n-1) + b(n-1), a(0) = 2, a(1) = 5, b(0) = 1
A295058: a(n) = 2*a(n-1) - b(n-1), a(0) = 3, a(1) = 5, b(0) = 1
A295059: a(n) = 2*a(n-1) + b(n-2), a(0) = 1, a(1) = 4, b(0) = 2, b(1) = 3
A295060: a(n) = 2*a(n-1) - b(n-2), a(0) = 3, a(1) = 5, b(0) = 1, b(1) = 2
A295061: a(n) = 4*a(n-1) + b(n-1), a(0) = 1, a(1) = 3, b(0) = 2
A295062: a(n) = 4*a(n-2) + b(n-2), a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4
A295063: a(n) = 4*a(n-2) + b(n-1) + b(n-2), a(0) = 1, a(1) = 3, b(0) = 2
A295064: a(n) = 8*a(n-3) + b(n-1), a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2
A295065: a(n) = 8*a(n-3) + b(n-2), a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2
A295066: a(n) = 2*a(n-2) + b(n-1), a(0) = 1, a(1) = 3, b(0) = 2
A295067: a(n) = 2*a(n-2) + b(n-2), a(0) = 1, a(1) = 3, b(0) = 2
A295068: a(n) = 2*a(n-2) - b(n-1) + n, a(0) = 3, a(1) = 4, b(0) = 1
A295069: a(n) = 2*a(n-2) - b(n-2) + n, a(0) = 3, a(1) = 4, b(0) = 1
A295070: a(n) = a(n-2) + b(n-1) + b(n-2), a(0) = 3, a(1) = 2, b(0) = 3
A295133: a(n) = 3*a(n-1) + b(n-1), a(0) = 1, a(1) = 2, b(0) = 3
A295134: a(n) = 3*a(n-1) + b(n-1) - 1, a(0) = 1, a(1) = 2, b(0) = 3
A295135: a(n) = 3*a(n-1) + b(n-1) - 2, a(0) = 1, a(1) = 2, b(0) = 3
A295136: a(n) = 3*a(n-1) + b(n-1) - 3, a(0) = 1, a(1) = 2, b(0) = 3
A295137: a(n) = 3*a(n-1) + b(n-1) - n, a(0) = 1, a(1) = 2, b(0) = 3
A295138: a(n) = 3*a(n-2) + b(n-1), a(0) = 1, a(1) = 2, b(0) = 3
A295139: a(n) = 3*a(n-1) + b(n-2), a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4
A295140: a(n) = 3*a(n-1) - b(n-2) + 4, a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4
A295141: a(n) = 2*a(n-1) + a(n-2) + b(n-2), a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4
A295142: a(n) = 2*a(n-1) + a(n-2) + b(n-2), a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4
A295143: a(n) = 2*a(n-1) + a(n-1) + b(n-1), a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4
A295144: a(n) = 2*a(n-1) + a(n-2) + b(n-1), a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4
A295145: a(n) = a(n-1) + 2*a(n-2) + b(n-2), a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4
A295146: a(n) = a(n-1) + 2*a(n-2) + b(n-2), a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4
A295147: a(n) = a(n-1) + 2*a(n-2) + b(n-1), a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4
A295148: a(n) = a(n-1) + 2*a(n-2) + b(n-1), a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4

Examples

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

Links

Crossrefs

Cf. A294860.

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 2; b[0] = 3;
a[n_] := a[n] = a[n - 1] + a[n - 2] + Sum[b[k], {k, 0, 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}]  (* A295053 *)
Table[b[n], {n, 0, 10}]

A295141 Solution of the complementary equation a(n) = 2*a(n-1) + a(n-2) + b(n-2), where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 2, 8, 22, 57, 142, 348, 847, 2052, 4962, 11988, 28951, 69904, 168774, 407468, 983727, 2374940, 5733626, 13842212, 33418071, 80678377, 194774849, 470228100
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, b(1) = 4
a(2) =2*a(1) + a(0) + b(0) = 8
Complement: (b(n)) = (3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, ...)

Links

Crossrefs

Cf. A295053, A295142, A295143, A295144.

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 2; b[0] = 3; b[1] = 4;
a[n_] := a[n] = 2 a[ n - 1] + 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}]  (* A295141 *)
Table[b[n], {n, 0, 10}]

Formula

a(n+1)/a(n) -> 1 + sqrt(2).

A295142 Solution of the complementary equation a(n) = 2*a(n-1) + 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, 9, 25, 64, 159, 389, 945, 2289, 5534, 13369, 32285, 77953, 188206, 454381, 1096985, 2648369, 6393742, 15435873, 37265509, 89966913, 217199358, 524365653, 1265930690
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) = 3, b(0) = 2, b(1) = 4
a(2) =2*a(1) + a(0) + b(0) = 9
Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, ...)

Links

Crossrefs

Cf. A295053, A295141, A295143, A295144.

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] = 2 a[ n - 1] + 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}]  (* A295142 *)
Table[b[n], {n, 0, 10}]

Formula

a(n+1)/a(n) -> 1 + sqrt(2).

A295143 Solution of the complementary equation a(n) = 2*a(n-1) + a(n-2) + b(n-1), where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 2, 9, 25, 65, 162, 397, 966, 2340, 5658, 13669, 33010, 79704, 192434, 464589, 1121630, 2707868, 6537386, 15782661, 38102730, 91988144, 222079042
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, b(1) = 4
a(2) =2*a(1) + a(0) + b(1) = 9
Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, ...)

Links

Crossrefs

Cf. A295053, A295141, A295142, A295144.

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 2; b[0] = 3; b[1] = 4;
a[n_] := a[n] = 2 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, 18}]  (* A295143 *)
Table[b[n], {n, 0, 10}]

Formula

a(n+1)/a(n) -> 1 + sqrt(2).
Showing 1-4 of 4 results.

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