A295359 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - 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, 14, 24, 41, 68, 112, 183, 298, 484, 786, 1275, 2064, 3342, 5409, 8754, 14166, 22923, 37092, 60019, 97116, 157138, 254257, 411398, 665658, 1077059, 1742720, 2819782, 4562505, 7382290, 11944798, 19327091, 31271892, 50598986

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) - 2* b(0) = 14
Complement: (b(n)) = (2, 4, 6, 7, 8, 9, 10, 11, 12, 13, 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] - 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}]   (* A295359 *)
v = Table[b[n], {n, 0, 10}]  (* complement *)

Crossrefs

Cf. A001622, A295357.

Sequence in context: A026645 A026667 A372436 * A104208 A378262 A364314

Adjacent sequences: A295356 A295357 A295358 * A295360 A295361 A295362

Keyword

nonn,easy

Author

Clark Kimberling, Nov 21 2017

Status

approved