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 13 results. Next

A296489 Decimal expansion of ratio-sum for A293358; see Comments.

Original entry on oeis.org

3, 5, 3, 5, 9, 1, 7, 3, 9, 3, 5, 1, 5, 2, 5, 9, 3, 5, 0, 7, 1, 6, 8, 0, 2, 2, 3, 2, 2, 4, 8, 5, 5, 0, 6, 7, 3, 3, 5, 5, 7, 5, 3, 4, 6, 2, 2, 7, 4, 4, 6, 8, 4, 2, 8, 0, 2, 2, 1, 0, 8, 9, 2, 8, 7, 4, 5, 0, 0, 5, 1, 2, 4, 2, 3, 1, 0, 0, 6, 9, 4, 5, 5, 7, 8, 9
Offset: 1

Views

Author

Clark Kimberling, Dec 19 2017

Keywords

Comments

Suppose that A = (a(n)), for n >= 0, is a sequence, and g is a real number such that a(n)/a(n-1) -> g. The ratio-sum for A is |a(1)/a(0) - g| + |a(2)/a(1) - g| + ..., assuming that this series converges. For A = A293358, we have g = (1 + sqrt(5))/2, the golden ratio (A001622). See the guide at A296469 for related sequences.

Examples

			ratio-sum = 3.535917393515259350716802232248550673355...

Crossrefs

Cf. A001622, A293358, A296284, A296490.

Programs

  • Mathematica
    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];
j = 1; While[j < 13, k = a[j] - j - 1;
While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];
Table[a[n], {n, 0, k}]; (* A293358 *)
g = GoldenRatio; s = N[Sum[- g + a[n]/a[n - 1], {n, 1, 1000}], 200]
Take[RealDigits[s, 10][[1]], 100]  (* A296489 *)

A296490 Decimal expansion of limiting power-ratio for A293358; see Comments.

Original entry on oeis.org

5, 7, 9, 2, 4, 9, 7, 4, 8, 4, 5, 2, 1, 0, 7, 5, 8, 9, 2, 7, 4, 7, 7, 3, 8, 0, 7, 5, 8, 8, 2, 8, 6, 7, 0, 1, 6, 8, 2, 2, 1, 4, 0, 8, 1, 7, 6, 5, 1, 7, 1, 8, 4, 0, 3, 6, 8, 7, 8, 9, 0, 1, 1, 6, 2, 1, 6, 4, 9, 0, 0, 2, 9, 3, 3, 6, 0, 6, 8, 1, 1, 4, 4, 6, 9, 4
Offset: 1

Views

Author

Clark Kimberling, Dec 19 2017

Keywords

Comments

Suppose that A = (a(n)), for n >= 0, is a sequence, and g is a real number such that a(n)/a(n-1) -> g. The limiting power-ratio for A is the limit as n->oo of a(n)/g^n, assuming that this limit exists. For A = A293358, we have g = (1 + sqrt(5))/2, the golden ratio (A001622). See the guide at A296469 for related sequences.

Examples

			limiting power-ratio = 5.792497484521075892747738075882867016822...

Crossrefs

Cf. A001622, A293358, A296284, A296489.

Programs

  • Mathematica
    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];
j = 1; While[j < 13, k = a[j] - j - 1;
While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];
Table[a[n], {n, 0, k}]; (* A293358 *)
z = 2000; g = GoldenRatio; h = Table[N[a[n]/g^n, z], {n, 0, z}];
StringJoin[StringTake[ToString[h[[z]]], 41], "..."]
Take[RealDigits[Last[h], 10][[1]], 120]   (* A296490 *)

A293076 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 6, 13, 24, 44, 76, 129, 215, 355, 582, 951, 1548, 2515, 4080, 6613, 10712, 17345, 28078, 45445, 73546, 119016, 192588, 311631, 504247, 815907, 1320184, 2136122, 3456338, 5592493, 9048865, 14641393, 23690294, 38331724, 62022056, 100353819, 162375915
Offset: 0

Views

Author

Clark Kimberling, Oct 28 2017

Keywords

Comments

The complementary sequences a() and b() are uniquely determined by the titular equation and initial values, which for each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:
A293076: a(n) = a(n-1) + a(n-2) + b(n-2)
A293316: a(n) = a(n-1) + a(n-2) + b(n-2) + 1
A293057: a(n) = a(n-1) + a(n-2) + b(n-2) + 2
A293058: a(n) = a(n-1) + a(n-2) + b(n-2) + 3
A293317: a(n) = a(n-1) + a(n-2) + b(n-2) - 1
A293349: a(n) = a(n-1) + a(n-2) + b(n-2) + n
A293350: a(n) = a(n-1) + a(n-2) + b(n-2) + 2*n
A293351: a(n) = a(n-1) + a(n-2) + b(n-2) + n - 1
A293357: a(n) = a(n-1) + a(n-2) + b(n-2) + n + 1
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(0) = 3 + 1 + 2 = 6;
a(3) = a(2) + a(1) + b(1) = 6 + 3 + 4 = 13.
Complement: (b(n)) = (2,4,5,7,8,9,10,11,12,14,...)

Links

Crossrefs

Cf. A001622 (golden ratio), A293358.

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 - 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}]  (* A293076 *)
Table[b[n], {n, 0, 10}]

Extensions

Comments corrected by Georg Fischer, Sep 23 2020

A296469 Decimal expansion of ratio-sum for A295862; see Comments.

Original entry on oeis.org

3, 8, 7, 0, 2, 3, 6, 0, 7, 9, 7, 9, 5, 9, 5, 9, 3, 2, 3, 2, 8, 2, 0, 5, 2, 3, 1, 1, 7, 8, 3, 9, 9, 5, 0, 1, 3, 8, 5, 6, 7, 3, 9, 8, 3, 0, 0, 9, 7, 2, 3, 1, 9, 9, 4, 3, 0, 1, 0, 8, 7, 6, 5, 5, 9, 5, 8, 0, 5, 4, 5, 4, 0, 6, 7, 3, 8, 5, 3, 9, 0, 5, 8, 8, 6, 2
Offset: 1

Views

Author

Clark Kimberling, Dec 18 2017

Keywords

Comments

Suppose that A = (a(n)), for n >= 0, is a sequence, and g is a real number such that a(n)/a(n-1) -> g. The ratio-sum for A is |a(1)/a(0) - g| + |a(2)/a(1) - g| + ..., assuming that this series converges. For A = A295862, we have g = (1 + sqrt(5))/2, the golden ratio (A001622). See A296425-A296434 for related ratio-sums and A296452-A296461 for related limiting power-ratios. Guide to more ratio-sums and limiting power-ratios:
****
Sequence A ratio-sum for A limiting power-ratio for A
A295862 A296469 A296470
A295947 A296471 A296472
A295948 A296473 A296474
A295949 A296475 A296476
A295950 A296477 A296478
A295951 A296479 A296480
A295952 A296481 A296482
A295953 A296483 A296848
A295960 A296485 A296486
A293076 A296487 A296488
A293358 A296489 A296490
A294170 A296491 A296492
A296555 A296493 A296494
A294414 A296495 A296496
A294541 A296497 A296498
A294546 A296499 A296500
A294552 A296501 A296494
A296776 A298171 A298172
A294553 A296503 A296504
A296556 A296565 A296566
A296557 A296567 A296568
A296558 A296569 A296570

Examples

			ratio-sum = 6.21032710946618494227967...

Crossrefs

Cf. A001622, A296284, A296470.

Programs

  • Mathematica
    a[0] = 1; a[1] = 3; b[0] = 2; b[1 ] = 4; b[2] = 5;
a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n];
j = 1; While[j < 13, k = a[j] - j - 1;
While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];
Table[a[n], {n, 0, k}]; (* A295862 *)
g = GoldenRatio; s = N[Sum[- g + a[n]/a[n - 1], {n, 1, 1000}], 200]
Take[RealDigits[s, 10][[1]], 100]  (* A296469 *)

A293765 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + 2, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 10, 20, 38, 67, 115, 193, 321, 528, 864, 1408, 2289, 3715, 6023, 9758, 15802, 25583, 41409, 67017, 108452, 175496, 283976, 459501, 743507, 1203039, 1946578, 3149650, 5096262, 8245947, 13342245, 21588229, 34930512, 56518780, 91449333, 147968155
Offset: 0

Views

Author

Clark Kimberling, Oct 29 2017

Keywords

Comments

The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:
Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio. See A293358 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) + b(1) + 2 = 10;
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, ...)

Links

Crossrefs

Cf. A001622 (golden ratio), A293765.

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_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 40}]  (* A293765 *)
Table[b[n], {n, 0, 10}]

A294476 Solution of the complementary equation a(n) = a(n-2) + b(n-1) + 1, where a(0) = 1, a(1) = 3, b(0) = 2.

Original entry on oeis.org

1, 3, 6, 9, 14, 18, 25, 30, 38, 44, 54, 61, 72, 81, 93, 103, 116, 127, 141, 154, 169, 183, 199, 215, 232, 249, 267, 285, 304, 323, 344, 364, 386, 407, 430, 453, 477, 501, 526, 551, 577, 603, 630, 657, 686, 714, 744, 773, 804, 834, 867, 898, 932, 964, 999
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. The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2:
A294476: a(n) = a(n-2) + b(n-1) + 1
A294477: a(n) = a(n-2) + b(n-1) + 2
A294478: a(n) = a(n-2) + b(n-1) + 3
A294479: a(n) = a(n-2) + b(n-1) + n
A294480: a(n) = a(n-2) + b(n-1) + 2n
A294481: a(n) = a(n-2) + b(n-1) + n - 1
A294482: a(n) = a(n-2) + b(n-1) + n + 1
For a(n-2) + b(n-1), with offset 1 instead of 0, see A022942.

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, 7, 8, 10, 11, 12, 13, 15,...)

Links

Crossrefs

Cf. A293076, A293765, A293358, A294414.

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] + 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}]  (* A294476 *)
Table[b[n], {n, 0, 10}]

A293406 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + 1, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 9, 18, 34, 60, 103, 174, 289, 476, 779, 1270, 2065, 3352, 5435, 8807, 14263, 23092, 37378, 60494, 97897, 158417, 256341, 414786, 671156, 1085972, 1757159, 2843163, 4600355, 7443552, 12043943, 19487532, 31531513, 51019084, 82550637, 133569762
Offset: 0

Views

Author

Clark Kimberling, Oct 29 2017

Keywords

Comments

The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:
Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio. See A293358 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) + b(1) + 1 = 8;
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, ...)

Links

Crossrefs

Cf. A001622 (golden ratio), A293076.

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] + 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}]  (* A293406 *)
Table[b[n], {n, 0, 10}]

A293766 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + 3, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 11, 22, 42, 74, 127, 213, 353, 581, 950, 1548, 2516, 4083, 6619, 10723, 17364, 28110, 45498, 73634, 119159, 192821, 312009, 504860, 816900, 1321792, 2138725, 3460551, 5599311, 9059898, 14659246, 23719182, 38378467, 62097689, 100476197, 162573928
Offset: 0

Views

Author

Clark Kimberling, Oct 29 2017

Keywords

Comments

The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:
Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio. See A293358 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) + b(1) + 3 = 11;
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, ...)

Links

Crossrefs

Cf. A001622 (golden ratio), A293765.

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] + 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}]  (* A293766 *)
Table[b[n], {n, 0, 10}]

A293767 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) - 1, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 7, 14, 26, 47, 81, 137, 228, 376, 616, 1006, 1637, 2659, 4313, 6990, 11322, 18332, 29675, 48029, 77727, 125780, 203533, 329340, 532901, 862270, 1395201, 2257502, 3652735, 5910270, 9563039, 15473344, 25036419, 40509800, 65546257, 106056096
Offset: 0

Views

Author

Clark Kimberling, Oct 29 2017

Keywords

Comments

The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:
Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio. See A293358 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) + b(1) - 1 = 7;
Complement: (b(n)) = (2, 4, 5, 6, 8, 9, 10, 11, 12, 13, 15, 16, 17, ...)

Links

Crossrefs

Cf. A001622 (golden ratio), A293765.

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] - 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}]  (* A293767 *)
Table[b[n], {n, 0, 10}]

A294365 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + n, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 10, 21, 41, 74, 129, 219, 367, 607, 997, 1629, 2653, 4311, 6995, 11339, 18369, 29745, 48154, 77941, 126139, 204126, 330313, 534489, 864854, 1399397, 2264307, 3663762, 5928129, 9591953, 15520146, 25112165, 40632379, 65744614, 106377065, 172121753
Offset: 0

Views

Author

Clark Kimberling, Oct 29 2017

Keywords

Comments

The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:
Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio. See A293358 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) + b(1) + 2 = 10;
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, ...)

Links

Crossrefs

Cf. A001622 (golden ratio), A293765.

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] + 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}]  (* A294365 *)
Table[b[n], {n, 0, 10}]
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