A299495 Solution b( ) of the complementary equation a(n) = b(n-1) + b(n-2) + b(n-3), where a(0) = 2, a(1) = 4, a(2) = 6; see Comments.

1, 3, 5, 7, 8, 10, 11, 12, 13, 14, 16, 17, 18, 19, 21, 22, 23, 24, 26, 27, 28, 30, 31, 32, 34, 35, 37, 38, 40, 41, 42, 44, 45, 46, 48, 49, 50, 52, 53, 55, 56, 57, 59, 60, 61, 63, 64, 65, 67, 68, 70, 71, 72, 74, 75, 76, 78, 79, 80, 82, 83, 84, 86, 87, 88, 90

Offset

0,2

Comments

From the Bode-Harborth-Kimberling link:

a(n) = b(n-1) + b(n-2) + b(n-3) for n > 2;

b(0) = least positive integer not in {a(0),a(1),a(2)};

b(n) = least positive integer not in {a(0),...,a(n),b(0),...,b(n-1)} for n > 1.

Note that (b(n)) is strictly increasing and is the complement of (a(n)).

See A022424 for a guide to related sequences.

Links

Table of n, a(n) for n=0..65.

J-P. Bode, H. Harborth, C. Kimberling, Complementary Fibonacci sequences, Fibonacci Quarterly 45 (2007), 254-264.

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] = 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}]]];
Table[a[n], {n, 0, 100}]    (* A299494 *)
Table[b[n], {n, 0, 100}]    (* A299495 *)

Crossrefs

Cf. A022424, A299494.

Sequence in context: A188569 A274140 A212294 * A186689 A186350 A360392

Adjacent sequences: A299492 A299493 A299494 * A299496 A299497 A299498

Keyword

nonn,easy

Author

Clark Kimberling, Feb 21 2018

Status

approved