A295615 Solution of the complementary equation a(n) = 2*a(n-1) - a(n-3) + b(n-1), where a(0) = 2, a(1) = 4, a(2) = 6, b(0) = 1, b(1) = 3, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.

2, 4, 6, 15, 33, 68, 130, 237, 417, 716, 1208, 2013, 3326, 5461, 8927, 14547, 23653, 38400, 62275, 100920, 163464, 264678, 428462, 693487, 1122324, 1816215, 2938973, 4755653, 7695123, 12451307, 20146996, 32598905, 52746540, 85346122, 138093378, 223440256

Offset

0,1

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

Mathematica

mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 2; a[1] = 4; a[2] = 6; b[0] = 1; b[1] = 3; b[2] = 5;
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}]  (* A295615 *)
Table[b[n], {n, 0, 20}]  (* complement *)

Crossrefs

Cf. A001622, A000045, A295613.

Sequence in context: A192536 A227902 A217525 * A127679 A049022 A209867

Adjacent sequences: A295612 A295613 A295614 * A295616 A295617 A295618

Keyword

nonn,easy

Author

Clark Kimberling, Nov 25 2017

Status

approved