A295362 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) - b(n-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, 8, 10, 14, 19, 25, 34, 49, 71, 106, 162, 253, 398, 632, 1010, 1621, 2610, 4208, 6793, 10975, 17741, 28688, 46400, 75058, 121428, 196454, 317848, 514267, 832079, 1346309, 2178350, 3524620, 5702930, 9227509, 14930397, 24157863

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..37.

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(2) + a(1) + b(2) - b(1) - b(0) = 8
Complement: (b(n)) = (2, 4, 6, 7, 9, 11, 12, 13, 15, 16, 17, ...)

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] - 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}]   (* A295362 *)
v = Table[b[n], {n, 0, 10}]  (* complement *)

Crossrefs

Cf. A001622, A295357.

Sequence in context: A253081 A088940 A088937 * A162383 A123327 A024679

Adjacent sequences: A295359 A295360 A295361 * A295363 A295364 A295365

Keyword

nonn,easy

Author

Clark Kimberling, Nov 21 2017

Status

approved