A295358 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.

1, 3, 5, 16, 30, 55, 95, 161, 268, 442, 724, 1181, 1921, 3120, 5061, 8201, 13283, 21506, 34812, 56342, 91179, 147547, 238753, 386328, 625110, 1011468, 1636610, 2648112, 4284756, 6932903, 11217695, 18150635, 29368368, 47519042, 76887450

Offset

0,2

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295357 for a guide to related sequences.

Links

Table of n, a(n) for n=0..34.

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

Formula

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).

Example

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) = 16
Complement: (b(n)) = (2, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, ...)

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}]   (* A295358 *)
v = Table[b[n], {n, 0, 10}]  (* complement *)

Crossrefs

Cf. A001622, A295357.

Sequence in context: A330055 A019096 A370604 * A369930 A077551 A106588

Adjacent sequences: A295355 A295356 A295357 * A295359 A295360 A295361

Keyword

nonn,easy

Author

Clark Kimberling, Nov 21 2017

Status

approved