A295619 -id:A295619 - cp's OEIS frontend

A293411 a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4.

1, 2, 3, 4, 7, 12, 19, 30, 49, 80, 129, 208, 337, 546, 883, 1428, 2311, 3740, 6051, 9790, 15841, 25632, 41473, 67104, 108577, 175682, 284259, 459940, 744199, 1204140, 1948339, 3152478, 5100817, 8253296, 13354113, 21607408, 34961521, 56568930, 91530451

Offset: 0

Clark Kimberling, Nov 25 2017

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622), so that this sequence has the growth rate of the Fibonacci numbers (A000045).

Clark Kimberling, Table of n, a(n) for n = 0..1999
Index entries for linear recurrences with constant coefficients, signature (1,0,1,1).

Cf. A001622, A000045, A295619, A295620, A061646, A079472.

LinearRecurrence[{1, 0, 1, 1}, {1, 2, 3, 4}, 100]

a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4.
G.f.: (1+x+x^2)/((1+x^2)*(1-x-x^2)).
From Greg Dresden, Aug 25 2021: (Start)
a(2*n) = a(2*n - 1) + a(2*n - 2),
a(2*n) = 2*F(n+1)^2 - (-1)^n = A061646(n+1),
a(2*n+1) = 2*F(n+1)*F(n+2) = A079472(n+2), for F(n) the Fibonacci numbers A000045. (End)
a(n) = round (2*A000032(n+2)/5). - R. J. Mathar, Jul 20 2025

A295620 Solution of the complementary equation a(n) = a(n-1) + 3a(n-2) -2a(n-3) - 2*a(n-4) + b(n-4), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 2, 3, 4, 12, 20, 49, 85, 177, 304, 578, 979, 1765, 2953, 5150, 8538, 14570, 23997, 40352, 66149, 110094, 179867, 297172, 484313, 795934, 1294823, 2119684, 3443689, 5621258, 9123343, 14860404, 24100573, 39192618, 63526879, 103182816, 167177109, 271286602

Offset: 0

Clark Kimberling, Nov 25 2017

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values.

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

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

Cf. A001622, A000045, A295619, A295621.

mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 2; a[2] = 3; a[3] = 4;
b[0] = 5; b[1] = 6; b[2] = 7; b[3] = 8;
a[n_] := a[n] = a[n - 1] + 3*a[n - 2] - 2*a[n - 3] - 2 a[n - 4] + b[n - 4];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
z = 36;  Table[a[n], {n, 0, z}]   (* A295620 *)
Table[b[n], {n, 0, 20}]  (*complement *)

A295621 Solution of the complementary equation a(n) = a(n-1) + 3a(n-2) -2a(n-3) - 2*a(n-4) + b(n-3), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 2, 3, 4, 13, 22, 55, 96, 201, 346, 659, 1117, 2015, 3372, 5882, 9752, 16643, 27411, 46093, 75559, 125754, 205448, 339432, 553177, 909097, 1478897, 2421000, 3933174, 6420218, 10419979, 16972319, 27525507, 44762106, 72554068, 117844772, 190931789, 309833797

Offset: 0

Clark Kimberling, Nov 25 2017

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values.

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

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

Cf. A001622, A000045, A295619, A295620.

mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 2; a[2] = 3; a[3] = 4;
b[0] = 5; b[1] = 6; b[2] = 7; b[3] = 8;
a[n_] := a[n] = a[n - 1] + 3*a[n - 2] - 2*a[n - 3] - 2 a[n - 4] + b[n - 3];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
z = 36;  Table[a[n], {n, 0, z}]   (* A295621 *)
Table[b[n], {n, 0, 20}]  (*complement *)

Typo in the definition corrected by Georg Fischer, Jun 08 2022

cp's OEIS Frontend

A293411 a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A295620 Solution of the complementary equation a(n) = a(n-1) + 3a(n-2) -2a(n-3) - 2*a(n-4) + b(n-4), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, and (a(n)) and (b(n)) are increasing complementary sequences.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A295621 Solution of the complementary equation a(n) = a(n-1) + 3a(n-2) -2a(n-3) - 2*a(n-4) + b(n-3), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, and (a(n)) and (b(n)) are increasing complementary sequences.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Extensions

cp's OEIS Frontend

A293411 a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A295620 Solution of the complementary equation a(n) = a(n-1) + 3*a(n-2) -2*a(n-3) - 2*a(n-4) + b(n-4), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, and (a(n)) and (b(n)) are increasing complementary sequences.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A295621 Solution of the complementary equation a(n) = a(n-1) + 3*a(n-2) -2*a(n-3) - 2*a(n-4) + b(n-3), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, and (a(n)) and (b(n)) are increasing complementary sequences.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Extensions

A295620 Solution of the complementary equation a(n) = a(n-1) + 3a(n-2) -2a(n-3) - 2*a(n-4) + b(n-4), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, and (a(n)) and (b(n)) are increasing complementary sequences.

A295621 Solution of the complementary equation a(n) = a(n-1) + 3a(n-2) -2a(n-3) - 2*a(n-4) + b(n-3), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, and (a(n)) and (b(n)) are increasing complementary sequences.