A295616 Solution of the complementary equation a(n) = 2*a(n-1) - a(n-3) + b(n-2), where a(0) = 1, a(1) = 2, a(2) = 3, b(0) = 4, b(1) = 5, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.

1, 2, 3, 10, 24, 52, 102, 189, 337, 584, 992, 1661, 2753, 4530, 7416, 12097, 19683, 31970, 51864, 84067, 136187, 220535, 357029, 577898, 935289, 1513578, 2449288, 3963318, 6413090, 10376925, 16790566, 27168077, 43959265, 71128001, 115087963, 186216700

Offset

0,2

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295613 for a guide to related sequences.

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

Links

Table of n, a(n) for n=0..35.

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

Example

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

Mathematica

mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 2; a[2] = 3; b[0] = 4; b[1] = 5; b[2] = 6;
a[n_] := a[n] = 2 a[n - 1] - a[n - 3] + 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, 30}]  (* A295616 *)
Table[b[n], {n, 0, 20}]  (* complement *)

Crossrefs

Cf. A001622, A000045, A295613.

Sequence in context: A320812 A162034 A105286 * A059929 A123029 A103018

Adjacent sequences: A295613 A295614 A295615 * A295617 A295618 A295619

Keyword

nonn,easy

Author

Clark Kimberling, Nov 25 2017

Status

approved