A299492 Solution a( ) of the complementary equation a(n) = b(n-1) + b(n-2) + b(n-3), where a(0) = 2, a(1) = 4, a(2) = 5; see Comments.

2, 4, 5, 10, 16, 21, 24, 28, 32, 36, 39, 42, 46, 50, 54, 57, 61, 65, 70, 74, 78, 82, 86, 90, 94, 98, 102, 106, 110, 115, 119, 124, 128, 132, 136, 140, 144, 148, 152, 156, 160, 164, 169, 173, 177, 181, 185, 189, 193, 197, 201, 204, 208, 212, 216, 220, 224

Offset

0,1

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..56.

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] = 5; b[0] = 1; b[1] = 3; b[2] = 6;
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}]    (* A299492 *)
Table[b[n], {n, 0, 100}]    (* A299493 *)

Crossrefs

Cf. A022424, A299493.

Sequence in context: A319909 A276281 A059994 * A253198 A138856 A333188

Adjacent sequences: A299489 A299490 A299491 * A299493 A299494 A299495

Keyword

nonn,easy

Author

Clark Kimberling, Feb 20 2018

Status

approved