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.

Previous Showing 31-40 of 40 results.

A294424 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, 5, 9, 14, 23, 38, 61, 99, 160, 260, 420, 680, 1100, 1780, 2880, 4660, 7540, 12201, 19741, 31942, 51683, 83625, 135308, 218933, 354241, 573174, 927415, 1500589, 2428004, 3928593, 6356597, 10285191, 16641788, 26926979, 43568767, 70495746, 114064513
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 = 5
Complement: (b(n)) = (2, 4, 6, 7, 9, 11, 12, 13, 15,...)

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}]  (* A294424 *)
Table[b[n], {n, 0, 10}]

A294425 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + 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, 9, 17, 32, 56, 96, 163, 270, 445, 728, 1187, 1930, 3133, 5082, 8234, 13336, 21591, 34949, 56563, 91536, 148124, 239686, 387837, 627551, 1015417, 1642998, 2658446, 4301478, 6959958, 11261471, 18221465, 29482973, 47704476, 77187488, 124892004, 202079533
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) + 2*b(1) - b(0) - 1 = 9
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 10, 11, 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] + 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}]  (* A294425 *)
Table[b[n], {n, 0, 10}]

A294426 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + 2*b(n-1) - b(n-2) - 2, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 8, 15, 28, 49, 86, 144, 240, 395, 647, 1055, 1718, 2789, 4524, 7331, 11874, 19225, 31120, 50367, 81510, 131901, 213436, 345363, 558828, 904220, 1463078, 2367329, 3830439, 6197801, 10028274, 16226110, 26254420, 42480567, 68735025, 111215631, 179950696
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) + 2*b(1) - b(0) - 2 = 8
Complement: (b(n)) = (2, 4, 5, 6, 7, 9, 10, 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] + 2*b[n - 1] - b[n - 2] - 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}]  (* A294426 *)
Table[b[n], {n, 0, 10}]

A294477 Solution of the complementary equation a(n) = a(n-2) + b(n-1) + 2, where a(0) = 1, a(1) = 3, b(0) = 2.

Original entry on oeis.org

1, 3, 7, 10, 15, 20, 26, 33, 40, 48, 56, 66, 75, 86, 96, 109, 120, 134, 146, 161, 175, 191, 206, 223, 239, 257, 275, 294, 313, 333, 353, 374, 396, 418, 441, 464, 488, 512, 537, 563, 589, 616, 643, 671, 699, 728, 758, 788, 819, 850, 882, 914, 947, 980, 1014
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 A294476 for a guide to related sequences.

Examples

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

Links

Crossrefs

Cf. A293076, A293765, A294476.

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] + 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}]  (* A294477 *)
Table[b[n], {n, 0, 10}]

A294478 Solution of the complementary equation a(n) = a(n-2) + b(n-1) + 3, where a(0) = 1, a(1) = 3, b(0) = 2.

Original entry on oeis.org

1, 3, 8, 11, 17, 21, 29, 34, 44, 50, 61, 68, 80, 89, 102, 112, 127, 138, 154, 166, 183, 196, 214, 229, 248, 264, 284, 302, 323, 342, 364, 384, 407, 428, 452, 474, 500, 523, 550, 574, 602, 628, 657, 684, 714, 742, 773, 802, 834, 864, 897, 929, 963, 996, 1031
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 A294476 for a guide to related sequences.

Examples

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

Links

Crossrefs

Cf. A293076, A293765, A294476.

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] + 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}]  (* A294478 *)
Table[b[n], {n, 0, 10}]

A294479 Solution of the complementary equation a(n) = a(n-2) + b(n-1) + n, where a(0) = 1, a(1) = 3, b(0) = 2.

Original entry on oeis.org

1, 3, 7, 11, 17, 24, 32, 41, 52, 63, 76, 89, 104, 120, 137, 155, 174, 194, 215, 238, 261, 286, 311, 338, 365, 394, 424, 455, 487, 520, 554, 589, 625, 662, 701, 740, 781, 822, 865, 908, 953, 998, 1045, 1092, 1142, 1191, 1243, 1294, 1348, 1401, 1457, 1512
Offset: 0

Views

Author

Clark Kimberling, Nov 01 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A294476 for a guide to related sequences.

Examples

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

Links

Crossrefs

Cf. A293076, A293765, A294476.

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] + 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}]  (* A294479 *)
Table[b[n], {n, 0, 10}]

A294480 Solution of the complementary equation a(n) = a(n-2) + b(n-1) + 2n, where a(0) = 1, a(1) = 3, b(0) = 2.

Original entry on oeis.org

1, 3, 9, 14, 23, 31, 43, 55, 70, 85, 103, 122, 143, 165, 189, 214, 241, 269, 299, 331, 364, 399, 435, 473, 512, 553, 596, 640, 686, 733, 782, 832, 884, 937, 992, 1048, 1106, 1166, 1227, 1290, 1354, 1420, 1487, 1556, 1626, 1698, 1771, 1846, 1923, 2001, 2081
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 A294476 for a guide to related sequences.

Examples

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

Links

Crossrefs

Cf. A293076, A293765, A294476.

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] + 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}]  (* A294480 *)
Table[b[n], {n, 0, 10}]

A294481 Solution of the complementary equation a(n) = a(n-2) + b(n-1) + n - 1, where a(0) = 1, a(1) = 3, b(0) = 2.

Original entry on oeis.org

1, 3, 6, 10, 16, 22, 30, 39, 49, 60, 72, 85, 100, 115, 132, 149, 168, 188, 209, 231, 254, 278, 303, 329, 357, 385, 415, 445, 477, 509, 543, 577, 614, 650, 689, 727, 768, 808, 851, 893, 938, 983, 1030, 1077, 1126, 1175, 1226, 1277, 1330, 1383, 1438, 1494
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 A294476 for a guide to related sequences.

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, 6, 7, 8, 10, 12, 13, 15, ...)

Links

Crossrefs

Cf. A293076, A293765, A294476.

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] + 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}]  (* A294481 *)
Table[b[n], {n, 0, 10}]

A294482 Solution of the complementary equation a(n) = a(n-2) + b(n-1) + n + 1, where a(0) = 1, a(1) = 3, b(0) = 2.

Original entry on oeis.org

1, 3, 8, 12, 19, 25, 35, 43, 55, 66, 80, 93, 109, 124, 142, 160, 180, 200, 222, 244, 269, 293, 320, 346, 375, 403, 434, 464, 497, 530, 565, 600, 637, 674, 713, 752, 794, 835, 879, 922, 968, 1013, 1061, 1108, 1158, 1207, 1259, 1311, 1365, 1419, 1475, 1531
Offset: 0

Views

Author

Clark Kimberling, Nov 03 2017

Keywords

Comments

The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A294476 for a guide to related sequences.

Examples

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

Links

Crossrefs

Cf. A293076, A293765, A294476.

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] + 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}]  (* A294482 *)
Table[b[n], {n, 0, 10}]

A294412 Solution of the complementary equation a(n) = a(n-1) + 2*b(n-2) + 1, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 8, 17, 28, 41, 56, 75, 96, 119, 144, 171, 200, 231, 264, 301, 340, 381, 424, 469, 516, 565, 616, 669, 724, 783, 844, 907, 972, 1039, 1108, 1179, 1252, 1327, 1404, 1483, 1564, 1649, 1736, 1825, 1916, 2009, 2104, 2201, 2300, 2401, 2504, 2609, 2716, 2825
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) + 2*b(0) + 1 = 8
Complement: (b(n)) = (2, 4, 5, 6, 7, 9, 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] + 2b[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}]  (* A294412 *)
Table[b[n], {n, 0, 10}]
Previous Showing 31-40 of 40 results.

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