A294414 -id:A294414 - cp's OEIS frontend

A296496 Decimal expansion of limiting power-ratio for A294414; see Comments.

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

Offset: 1

Clark Kimberling, Dec 20 2017

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 = A294414, we have g = (1 + sqrt(5))/2, the golden ratio (A001622). See the guide at A296469 for related sequences.

			limiting power-ratio = 8.814104632202565527925188322585412678508...

Cf. A001622, A294414, A296284, A296495.

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];
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}]; (* A294414 *)
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]   (* A296496 *)

A296495 Decimal expansion of ratio-sum for A294414; see Comments.

Original entry on oeis.org

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

Offset: 1

Clark Kimberling, Dec 20 2017

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 = A294414, we have g = (1 + sqrt(5))/2, the golden ratio (A001622). See the guide at A296469 for related sequences.

			4.585197174184565401988869057817650746901...

Cf. A001622, A294414, A296284, A296496.

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];
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}]; (* A294414 *)
g = GoldenRatio; s = N[Sum[- g + a[n]/a[n - 1], {n, 1, 1000}], 200]
Take[RealDigits[s, 10][[1]], 100]  (* A296495 *)

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

Clark Kimberling, Dec 18 2017

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

			ratio-sum = 6.21032710946618494227967...

Cf. A001622, A296284, A296470.

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 *)

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

Clark Kimberling, Nov 01 2017

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.

			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,...)

Clark Kimberling, Table of n, a(n) for n = 0..1000
Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.

Cf. A293076, A293765, A293358, A294414.

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}]

A294413 Solution of the complementary equation a(n) = a(n-1) + a(n-2) - b(n-1) + 6, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 6, 10, 15, 23, 35, 53, 82, 128, 202, 320, 511, 819, 1317, 2122, 3424, 5530, 8936, 14447, 23363, 37789, 61130, 98896, 160002, 258873, 418849, 677695, 1096516, 1774181, 2870666, 4644815, 7515448, 12160229, 19675642, 31835835

Offset: 0

Clark Kimberling, Oct 31 2017

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.

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

Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.

Cf. A293076, A293765, A294414.

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] + 6;
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 40}]  (* A294413 *)
Table[b[n], {n, 0, 10}]

A294415 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, 11, 24, 47, 85, 148, 251, 419, 693, 1138, 1859, 3027, 4918, 7979, 12933, 20950, 33923, 54915, 88882, 143843, 232774, 376669, 609497, 986222, 1595777, 2582059, 4177898, 6760021, 10937985, 17698074, 28636129, 46334275, 74970478, 121304829, 196275385

Offset: 0

Clark Kimberling, Oct 31 2017

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.

			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 = 11
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14,...)

Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.

Cf. A293076, A293765, A294414.

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}]  (* A294415 *)
Table[b[n], {n, 0, 10}]

A294416 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + n, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 12, 27, 54, 99, 174, 297, 498, 825, 1357, 2220, 3618, 5882, 9547, 15479, 25079, 40614, 65752, 106428, 172245, 278741, 451057, 729872, 1181007, 1910961, 3092053, 5003102, 8095246, 13098442, 21193785, 34292327, 55486215, 89778648, 145264972, 235043732

Offset: 0

Clark Kimberling, Oct 31 2017

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.

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

Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.

Cf. A293076, A293765, A294414.

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] + 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}]  (* A294416 *)
Table[b[n], {n, 0, 10}]

A294417 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - n, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 8, 17, 32, 57, 99, 168, 280, 462, 757, 1235, 2009, 3262, 5291, 8575, 13889, 22488, 36402, 58916, 95345, 154289, 249663, 403982, 653676, 1057690, 1711399, 2769123, 4480558, 7249719, 11730316, 18980075, 30710432, 49690549, 80401024, 130091617

Offset: 0

Clark Kimberling, Oct 31 2017

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.

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

Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.

Cf. A293076, A293765, A294414.

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] - 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}]  (* A294417 *)
Table[b[n], {n, 0, 10}]

A294418 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + 2*b(n-2), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 12, 28, 56, 103, 181, 309, 518, 858, 1411, 2309, 3763, 6118, 9930, 16100, 26085, 42243, 68389, 110696, 179152, 289918, 469143, 759137, 1228359, 1987579, 3216026, 5203696, 8419816, 13623609, 22043525, 35667237, 57710868, 93378214, 151089194, 244467523

Offset: 0

Clark Kimberling, Oct 31 2017

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.

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

Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.

Cf. A293076, A293765, A294414.

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 - 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}]  (* A294418 *)
Table[b[n], {n, 0, 10}]

A294419 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + 2b(n-1) + 2b(n-2), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 16, 37, 75, 138, 243, 415, 696, 1153, 1895, 3098, 5047, 8203, 13314, 21587, 34975, 56640, 91697, 148423, 240210, 388727, 629035, 1017864, 1647005, 2664979, 4312098, 6977195, 11289415, 18266736, 29556281, 47823151, 77379570, 125202863, 202582581

Offset: 0

Clark Kimberling, Oct 31 2017

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.

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

Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.

Cf. A293076, A293765, A294414.

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] + 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}]  (* A294419 *)
Table[b[n], {n, 0, 10}]

cp's OEIS Frontend

A296496 Decimal expansion of limiting power-ratio for A294414; see Comments.

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A296495 Decimal expansion of ratio-sum for A294414; see Comments.

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

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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.

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A294413 Solution of the complementary equation a(n) = a(n-1) + a(n-2) - b(n-1) + 6, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

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A294415 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.

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A294416 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + n, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

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A294417 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - n, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

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A294418 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + 2*b(n-2), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

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A294419 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + 2b(n-1) + 2b(n-2), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.