A295614 Solution of the complementary equation a(n) = 2*a(n-1) - a(n-3) + b(n-1), 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, 15, 34, 71, 136, 248, 436, 748, 1261, 2100, 3468, 5692, 9302, 15155, 24638, 39995, 64857, 105099, 170227, 275622, 446171, 722142, 1168690, 1891238, 3060364, 4952069, 8012932, 12965533, 20979032, 33945168, 54924840, 88870686, 143796243, 232667686

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

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] = 2 a[n - 1] - a[n - 3] + 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, 30}]  (* A295614 *)
Table[b[n], {n, 0, 20}]  (* complement *)

Crossrefs

Cf. A001622, A000045, A295613.

Sequence in context: A148500 A148501 A148502 * A089485 A279684 A146212

Adjacent sequences: A295611 A295612 A295613 * A295615 A295616 A295617

Keyword

nonn,easy

Author

Clark Kimberling, Nov 25 2017

Status

approved