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 51 results. Next

A296487 Decimal expansion of ratio-sum for A293076; see Comments.

Original entry on oeis.org

3, 0, 9, 2, 2, 6, 2, 2, 8, 5, 7, 7, 3, 1, 0, 6, 3, 3, 0, 1, 8, 3, 5, 3, 4, 6, 5, 5, 2, 0, 2, 7, 1, 6, 1, 6, 2, 4, 2, 5, 9, 4, 5, 8, 5, 3, 6, 9, 4, 2, 4, 6, 2, 4, 5, 5, 0, 6, 7, 2, 9, 0, 8, 0, 6, 9, 5, 8, 3, 5, 9, 6, 3, 1, 8, 2, 6, 8, 5, 5, 6, 2, 4, 7, 7, 3
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 = A293076, we have g = (1 + sqrt(5))/2, the golden ratio (A001622). See the guide at A296469 for related sequences.

Examples

			ratio-sum = 3.092262285773106330183534655202716162425...

Crossrefs

Cf. A001622, A293076, A296284, A296488.

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 - 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}]; (* A293076 *)
g = GoldenRatio; s = N[Sum[- g + a[n]/a[n - 1], {n, 1, 1000}], 200]
Take[RealDigits[s, 10][[1]], 100]  (* A296487 *)

A296488 Decimal expansion of limiting power-ratio for A293076; see Comments.

Original entry on oeis.org

4, 8, 6, 3, 6, 9, 8, 8, 6, 8, 1, 5, 6, 0, 7, 9, 1, 9, 5, 8, 5, 9, 8, 8, 8, 7, 5, 2, 1, 4, 9, 6, 5, 7, 1, 9, 8, 7, 1, 7, 4, 9, 0, 9, 2, 2, 2, 5, 6, 9, 4, 8, 8, 2, 3, 8, 9, 7, 6, 2, 2, 3, 2, 9, 1, 6, 7, 9, 6, 4, 4, 5, 0, 1, 6, 1, 7, 1, 3, 3, 9, 0, 8, 6, 3, 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 limiting power-ratio for A is the limit as n->oo of a(n)/g^n, assuming that this limit exists. For A = A293076, we have g = (1 + sqrt(5))/2, the golden ratio (A001622). See the guide at A296469 for related sequences.

Examples

			limiting power-ratio = 4.863698868156079195859888752149657198717...

Crossrefs

Cf. A001622, A293076, A296284, A296487.

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 - 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}]; (* A293076 *)
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]   (* A296488 *)

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 *)

A294532 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2), where a(0) = 1, a(1) = 2, b(0) = 3.

Original entry on oeis.org

1, 2, 6, 12, 23, 42, 73, 124, 207, 342, 562, 918, 1495, 2429, 3941, 6388, 10348, 16756, 27125, 43903, 71052, 114980, 186058, 301065, 487151, 788245, 1275426, 2063702, 3339160, 5402895, 8742089, 14145019, 22887144, 37032200, 59919382, 96951621, 156871043
Offset: 0

Views

Author

Clark Kimberling, Nov 03 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values, which, for the sequences in the following guide, are a(0) = 1, a(1) = 2, b(0) = 3:
a(n) = a(n-1) + a(n-2) + b(n-2) A294532
a(n) = a(n-1) + a(n-2) + b(n-2) + 1 A294533
a(n) = a(n-1) + a(n-2) + b(n-2) + 2 A294534
a(n) = a(n-1) + a(n-2) + b(n-2) + 3 A294535
a(n) = a(n-1) + a(n-2) + b(n-2) - 1 A294536
a(n) = a(n-1) + a(n-2) + b(n-2) + n A294537
a(n) = a(n-1) + a(n-2) + b(n-2) + 2n A294538
a(n) = a(n-1) + a(n-2) + b(n-2) + n - 1 A294539
a(n) = a(n-1) + a(n-2) + b(n-2) + 2n - 1 A294540
a(n) = a(n-1) + a(n-2) + b(n-1) A294541
a(n) = a(n-1) + a(n-2) + b(n-1) + 1 A294542
a(n) = a(n-1) + a(n-2) + b(n-1) + 2 A294543
a(n) = a(n-1) + a(n-2) + b(n-1) + 3 A294544
a(n) = a(n-1) + a(n-2) + b(n-1) - 1 A294545
a(n) = a(n-1) + a(n-2) + b(n-1) + n A294546
a(n) = a(n-1) + a(n-2) + b(n-1) + 2n A294547
a(n) = a(n-1) + a(n-2) + b(n-1) + n - 1 A294548
a(n) = a(n-1) + a(n-2) + b(n-1) + n + 1 A294549
a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) A294550
a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + 1 A294551
a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + n A294552
a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - n A294553
a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + 2 A294554
a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + 3 A294555
a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + n + 1 A294556
a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + n - 1 A294557
a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + 2n A294558
a(n) = a(n-1) + a(n-2) + b(n-1) + 2*b(n-2) A294559
a(n) = a(n-1) + a(n-2) + 2*b(n-1) + 2*b(n-2) A294560
a(n) = a(n-1) + a(n-2) + 2*b(n-1) + b(n-2) A294561
a(n) = a(n-1) + a(n-2) + b(n-1) - b(n-2) + 1 A294562
a(n) = a(n-1) + a(n-2) + b(n-1) - b(n-2) + n A294563
a(n) = a(n-1) + a(n-2) + 2*b(n-1) - b(n-2) - 1 A294564
a(n) = a(n-1) + a(n-2) + 2*b(n-1) - b(n-2) - 3 A294565
Conjecture: for every sequence listed here, a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).

Examples

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

Links

Crossrefs

Cf. A001622, A293076, A294413.

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 - 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}]  (* A294532 *)
Table[b[n], {n, 0, 10}]

A294414 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.

Original entry on oeis.org

1, 3, 10, 22, 43, 78, 136, 231, 387, 641, 1053, 1721, 2803, 4555, 7391, 11981, 19409, 31429, 50879, 82352, 133278, 215679, 349008, 564740, 913803, 1478600, 2392462, 3871123, 6263648, 10134836, 16398551, 26533456, 42932078, 69465607, 112397760, 181863444
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. The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:
A294413: a(n) = a(n-1) + a(n-2) - b(n-1) + 6
A294414: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2)
A294415: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + 1
A294416: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + n
A294417: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - n
A294418: a(n) = a(n-1) + a(n-2) + b(n-1) + 2*b(n-2)
A294419: a(n) = a(n-1) + a(n-2) + 2*b(n-1) + 2*b(n-2)
A294420: a(n) = a(n-1) + a(n-2) + 2*b(n-1) + b(n-2)
A294421: a(n) = a(n-1) + a(n-2) + 2*b(n-1) - b(n-2)
A294422: a(n) = a(n-1) + a(n-2) + b(n-1) - b(n-2) + 1
A294423: a(n) = a(n-1) + a(n-2) + b(n-1) - b(n-2) + n
A294424: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - 1
A294425: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - 2
A294426: a(n) = a(n-1) + 2*a(n-2) + b(n-1) + b(n-2) - 3
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) = 6
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 11, 12, 13, ...)

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] + 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}]]];
Table[a[n], {n, 0, 40}]  (* A294414 *)
Table[b[n], {n, 0, 10}]

Extensions

Definition corrected by Georg Fischer, Sep 27 2020

A293358 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 8, 16, 30, 53, 92, 155, 258, 425, 696, 1135, 1846, 2998, 4862, 7879, 12761, 20661, 33444, 54128, 87596, 141749, 229371, 371147, 600546, 971722, 1572299, 2544053, 4116385, 6660472, 10776892, 17437400, 28214329, 45651767, 73866135, 119517942
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:
A293358: a(n) = a(n-1) + a(n-2) + b(n-1)
A293406: a(n) = a(n-1) + a(n-2) + b(n-1) + 1
A293765: a(n) = a(n-1) + a(n-2) + b(n-1) + 2
A293766: a(n) = a(n-1) + a(n-2) + b(n-1) + 3
A293767: a(n) = a(n-1) + a(n-2) + b(n-1) - 1
A294365: a(n) = a(n-1) + a(n-2) + b(n-1) + n
A294366: a(n) = a(n-1) + a(n-2) + b(n-1) + 2n
A294367: a(n) = a(n-1) + a(n-2) + b(n-1) + n - 1
A294368: a(n) = a(n-1) + a(n-2) + b(n-1) + 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(1) = 8;
Complement: (b(n)) = (2, 4, 5, 6, 7, 9, 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];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 40}]  (* A293358 *)
Table[b[n], {n, 0, 10}]

A295357 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, 20, 42, 83, 149, 259, 438, 730, 1204, 1973, 3219, 5237, 8504, 13792, 22350, 36200, 58612, 94878, 153559, 248509, 402143, 650730, 1052954, 1703768, 2756809, 4460667, 7217569, 11678332, 18896000, 30574434, 49470539
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. Guide to related sequences:
***** Part 1: initial values are a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6
A295357: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + b(n-3)
A295358: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - b(n-3)
A295359: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - 2*b(n-3)
A295360: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - 3*b(n-3)
A295361: a(n) = a(n-1) + a(n-2) + b(n-1) + 2*b(n-2) - 3*b(n-3)
A295362: a(n) = a(n-1) + a(n-2) + b(n-1) - b(n-2) - b(n-3)
***** Part 2: initial values as shown
A295363: a(n) = a(n-1) + a(n-2) + b(n-1)*b(n-2); a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4
A295364: a(n) = a(n-1) + a(n-2) + b(n-1)*b(n-2); a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4
A295365: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + b(n-3); a(0) = 1, a(1) = 2, a(2) = 3, b(0) = 4, b(1) = 5, b(2) = 6
A295366: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - b(n-3); a(0) = 1, a(1) = 2, a(2) = 3, b(0) = 4, b(1) = 5, b(2) = 6
A295367: a(n) = a(n-1) + a(n-2) + b(n-1)*b(n-2); a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4
For all of these sequences, a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).

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) = 20
Complement: (b(n)) = (2, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, ...)

Links

Crossrefs

Cf. A001622, A293076, A294532.

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}]   (* A295357 *)
v = Table[b[n], {n, 0, 10}]  (* complement *)

A022940 a(n) = a(n-1) + b(n-2) for n >= 3, a( ) increasing, given a(1) = 1, a(2) = 3; where b( ) is complement of a( ).

Original entry on oeis.org

1, 3, 5, 9, 15, 22, 30, 40, 51, 63, 76, 90, 106, 123, 141, 160, 180, 201, 224, 248, 273, 299, 326, 354, 383, 414, 446, 479, 513, 548, 584, 621, 659, 698, 739, 781, 824, 868, 913, 959, 1006, 1054, 1103, 1153, 1205, 1258, 1312, 1367, 1423, 1480
Offset: 1

Views

Author

Clark Kimberling

Keywords

Comments

From Clark Kimberling, Oct 30 2017: (Start)
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:
here: a(n) = a(n-1) + b(n-2) [with a different offset]
A294397: a(n) = a(n-1) + b(n-2) + 1;
A294398: a(n) = a(n-1) + b(n-2) + 2;
A294399: a(n) = a(n-1) + b(n-2) + 3;
A294400: a(n) = a(n-1) + b(n-2) + n;
A294401: a(n) = a(n-1) + b(n-2) + 2*n.
(End)

Examples

			a(1) = 1, a(2) = 3, b(1) = 2, b(2) = 4, so that a(3) = a(2) + a(1) + b(2) = 5.
Complement: {b(n)} = {2, 4, 6, 7, 8, 10, 11, 12, 13, 14, 16, ...}

Links

Crossrefs

Cf. A005228 and references therein.
Cf. A293076, A293765, A294381.

Programs

  • Mathematica
    Fold[Append[#1, #1[[-1]] + Complement[Range[Max@#1 + 1], #1][[#2]]] &, {1, 3}, Range[50]] (* Ivan Neretin, Apr 04 2016 *)

A294381 Solution of the complementary equation a(n) = a(n-1)*b(n-2), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

Original entry on oeis.org

1, 3, 6, 24, 120, 840, 6720, 60480, 604800, 6652800, 79833600, 1037836800, 14529715200, 217945728000, 3487131648000, 59281238016000, 1067062284288000, 20274183401472000, 405483668029440000, 8515157028618240000, 187333454629601280000, 4308669456480829440000, 107716736412020736000000
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:
A294381: a(n) = a(n-1)*b(n-2)
A294382: a(n) = a(n-1)*b(n-2) - 1
A294383: a(n) = a(n-1)*b(n-2) + 1
A294384: a(n) = a(n-1)*b(n-2) - n
A294385: a(n) = a(n-1)*b(n-2) + n

Examples

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

Links

Crossrefs

Cf. A293076, 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]*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}]  (* A294381 *)
Table[b[n], {n, 0, 10}]

Extensions

More terms from Jack W Grahl, Apr 26 2018

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}]
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