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 10 results.

A295367 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) = 2, b(0) = 3, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 2, 15, 37, 82, 161, 299, 532, 921, 1563, 2616, 4335, 7133, 11692, 19097, 31095, 50534, 82009, 132963, 215434, 348903, 564889, 914392, 1479931, 2395025, 3875712, 6271549, 10148131, 16420610, 26569733, 42991399, 69562254, 112554843
Offset: 0

Views

Author

Clark Kimberling, Nov 21 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295357 for a guide to related sequences.

Examples

			a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, so that
b(2) = 5 (least "new number")
a(2) = a(1) + a(0) + b(1)*b(0) = 15
Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, 17, ...)
		

Crossrefs

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] = 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}]]];
    z = 32; u = Table[a[n], {n, 0, z}]   (* A295367 *)
    v = Table[b[n], {n, 0, 10}]  (* complement *)

Formula

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).

A295363 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, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 3, 12, 35, 77, 154, 287, 513, 890, 1513, 2546, 4241, 6997, 11478, 18747, 30531, 49620, 80531, 130571, 211564, 342641, 554757, 897998, 1453405, 2352105, 3806266, 6159183, 9966319, 16126432, 26093743, 42221231
Offset: 0

Views

Author

Clark Kimberling, Nov 21 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295357 for a guide to related sequences.

Examples

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

Crossrefs

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}]]];
    z = 32; u = Table[a[n], {n, 0, z}]   (* A295363 *)
    v = Table[b[n], {n, 0, 10}]  (* complement *)

Formula

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).

A295358 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - b(n-3), where a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 3, 5, 16, 30, 55, 95, 161, 268, 442, 724, 1181, 1921, 3120, 5061, 8201, 13283, 21506, 34812, 56342, 91179, 147547, 238753, 386328, 625110, 1011468, 1636610, 2648112, 4284756, 6932903, 11217695, 18150635, 29368368, 47519042, 76887450
Offset: 0

Views

Author

Clark Kimberling, Nov 21 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295357 for a guide to related sequences.

Examples

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

Crossrefs

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
    a[0] = 1; a[1] = 3; a[2] = 5; b[0] = 2; b[1] = 4; b[2] = 6;
    a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + b[n - 2] - b[n - 3];
    b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
    z = 32; u = Table[a[n], {n, 0, z}]   (* A295358 *)
    v = Table[b[n], {n, 0, 10}]  (* complement *)

Formula

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).

A295359 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - 2*b(n-3), where a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 3, 5, 14, 24, 41, 68, 112, 183, 298, 484, 786, 1275, 2064, 3342, 5409, 8754, 14166, 22923, 37092, 60019, 97116, 157138, 254257, 411398, 665658, 1077059, 1742720, 2819782, 4562505, 7382290, 11944798, 19327091, 31271892, 50598986
Offset: 0

Views

Author

Clark Kimberling, Nov 21 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295357 for a guide to related sequences.

Examples

			a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, so that
b(3) = 7 (least "new number")
a(3) = a(1) + a(0) + b(2) + b(1) - 2* b(0) = 14
Complement: (b(n)) = (2, 4, 6, 7, 8, 9, 10, 11, 12, 13, 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; b[1] = 4; b[2] = 6;
    a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + b[n - 2] - 2*b[n - 3];
    b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
    z = 32; u = Table[a[n], {n, 0, z}]   (* A295359 *)
    v = Table[b[n], {n, 0, 10}]  (* complement *)

Formula

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).

A295360 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - 3*b(n-3), where a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 3, 5, 12, 18, 27, 41, 63, 98, 155, 247, 392, 628, 1008, 1624, 2620, 4228, 6831, 11041, 17853, 28874, 46706, 75559, 122244, 197778, 319996, 517747, 837715, 1355433, 2193118, 3548520, 5741606, 9290093, 15031665, 24321723
Offset: 0

Views

Author

Clark Kimberling, Nov 21 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295357 for a guide to related sequences.

Examples

			a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, so that
b(3) = 7 (least "new number")
a(3) = a(1) + a(0) + b(2) + b(1) - 3* b(0) = 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; b[1] = 4; b[2] = 6;
    a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + b[n - 2] - 3*b[n - 3];
    b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
    z = 32; u = Table[a[n], {n, 0, z}]   (* A295360 *)
    v = Table[b[n], {n, 0, 10}]  (* complement *)

Formula

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).

A295361 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + 2*b(n-2) - b(n-3), where a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 3, 5, 20, 40, 76, 134, 230, 386, 640, 1052, 1720, 2802, 4554, 7390, 11980, 19408, 31429, 50882, 82357, 133287, 215694, 349033, 564781, 913870, 1478709, 2392639, 3871410, 6264113, 10135589, 16399770, 26535429, 42935271, 69470774, 112406121
Offset: 0

Views

Author

Clark Kimberling, Nov 21 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295357 for a guide to related sequences.

Examples

			a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, so that
b(3) = 7 (least "new number")
a(3) = a(1) + a(0) + b(2) + 2*b(1) - b(0) = 20
Complement: (b(n)) = (2, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, ...)
		

Crossrefs

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
    a[0] = 1; a[1] = 3; a[2] = 5; b[0] = 2; b[1] = 4; b[2] = 6;
    a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + 2*b[n - 2] - b[n - 3];
    b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
    z = 32; u = Table[a[n], {n, 0, z}]   (* A295361 *)
    v = Table[b[n], {n, 0, 10}]  (* complement *)

Formula

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).

A295362 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) - b(n-2) - b(n-3), where a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 3, 5, 8, 10, 14, 19, 25, 34, 49, 71, 106, 162, 253, 398, 632, 1010, 1621, 2610, 4208, 6793, 10975, 17741, 28688, 46400, 75058, 121428, 196454, 317848, 514267, 832079, 1346309, 2178350, 3524620, 5702930, 9227509, 14930397, 24157863
Offset: 0

Views

Author

Clark Kimberling, Nov 21 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295357 for a guide to related sequences.

Examples

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

Crossrefs

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
    a[0] = 1; a[1] = 3; a[2] = 5; b[0] = 2; b[1] = 4; b[2] = 6;
    a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] - b[n - 2] - b[n - 3];
    b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
    z = 32; u = Table[a[n], {n, 0, z}]   (* A295362 *)
    v = Table[b[n], {n, 0, 10}]  (* complement *)

Formula

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).

A295364 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, a[2] = 5, b(0) = 2, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 3, 5, 32, 79, 167, 318, 575, 1003, 1710, 2869, 4761, 7840, 12841, 20953, 34100, 55395, 89875, 145690, 236027, 382223, 618802, 1001625, 1621077, 2623404, 4245237, 6869453, 11115560, 17985943, 29102526, 47089591
Offset: 0

Views

Author

Clark Kimberling, Nov 21 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295357 for a guide to related sequences.

Examples

			a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, so that
b(2) = 6 (least "new number")
a(3) = a(2) + a(1) + b(2)*b(1) = 32
Complement: (b(n)) = (2, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, ...)
		

Crossrefs

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
    a[0] = 1; a[1] = 3; a[2] = 5; 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}]]];
    z = 32; u = Table[a[n], {n, 0, z}]   (* A295364 *)
    v = Table[b[n], {n, 0, 10}]  (* complement *)

Formula

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).

A295365 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + b(n-3), where a(0) = 1, a(1) = 2, a[2] = 3, b(0) = 4, b(1) = 5, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 2, 3, 20, 41, 82, 147, 256, 433, 722, 1191, 1952, 3185, 5182, 8415, 13648, 22117, 35823, 58002, 93891, 151962, 245925, 397962, 643965, 1042008, 1686057, 2728152, 4414299, 7142544, 11556939, 18699582
Offset: 0

Views

Author

Clark Kimberling, Nov 21 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295357 for a guide to related sequences.

Examples

			a(0) = 1, a(1) = 2, a[2] = 3, b(0) = 4, b(1) = 5, b(2) = 6, so that
b(2) = 7 (least "new number")
a(3) = a(2) + a(1) + b(2) + b(1) + b(0) = 20
Complement: (b(n)) = (4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, ...)
		

Crossrefs

Programs

  • Mathematica
    mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
    a[0] = 1; a[1] = 2; a[2] = 3; b[0] = 4; b[1] = 5; b[2]=6;
    a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + b[n - 2] + b[n - 3];
    b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
    z = 32; u = Table[a[n], {n, 0, z}]   (* A295365 *)
    v = Table[b[n], {n, 0, 10}]  (* complement *)

Formula

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).

A295366 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - b(n-3), where a(0) = 1, a(1) = 2, a[2] = 3, b(0) = 4, b(1) = 5, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 2, 3, 12, 23, 44, 77, 132, 221, 367, 604, 987, 1608, 2613, 4240, 6873, 11134, 18029, 29186, 47240, 76453, 123720, 200201, 323950, 524181, 848162, 1372375, 2220570, 3592979, 5813584, 9406599, 15220220, 24626857
Offset: 0

Views

Author

Clark Kimberling, Nov 22 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295357 for a guide to related sequences.

Examples

			a(0) = 1, a(1) = 2, a[2] = 3, b(0) = 4, b(1) = 5, b(2) = 6, so that
b(2) = 7 (least "new number")
a(3) = a(2) + a(1) + b(2) + b (1) - b(0) = 12
Complement: (b(n)) = (4, 5, 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] = 2; a[2] = 3; b[0] = 4; b[1] = 5; b[2]=6;
    a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + b[n - 2] - b[n - 3];
    b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
    z = 32; u = Table[a[n], {n, 0, z}]   (* A295366 *)
    v = Table[b[n], {n, 0, 10}]  (* complement *)

Formula

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).
Showing 1-10 of 10 results.