A295360 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - 3*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, 12, 18, 27, 41, 63, 98, 155, 247, 392, 628, 1008, 1624, 2620, 4228, 6831, 11041, 17853, 28874, 46706, 75559, 122244, 197778, 319996, 517747, 837715, 1355433, 2193118, 3548520, 5741606, 9290093, 15031665, 24321723

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) - 3* b(0) = 12
Complement: (b(n)) = (2, 4, 6, 7, 8, 9, 10, 11, 13, 14, 15, ...)

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] - 3*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}]   (* A295360 *)
v = Table[b[n], {n, 0, 10}]  (* complement *)

Crossrefs

Cf. A001622, A295357.

Sequence in context: A305552 A226652 A024696 * A197988 A025088 A279784

Adjacent sequences: A295357 A295358 A295359 * A295361 A295362 A295363

Keyword

nonn,easy

Author

Clark Kimberling, Nov 21 2017

Status

approved