A299491 Solution b( ) of the complementary equation a(n) = b(n-1) + b(n-2) + b(n-3), where a(0) = 1, a(1) = 3, a(2) = 5; see Comments.

2, 4, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 18, 19, 20, 22, 23, 25, 26, 28, 29, 31, 32, 33, 35, 36, 37, 39, 40, 41, 43, 44, 46, 47, 48, 50, 51, 52, 54, 55, 56, 58, 59, 60, 62, 63, 64, 66, 67, 68, 69, 71, 72, 73, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 89

Offset

0,1

Comments

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

Clark Kimberling, Table of n, a(n) for n = 0..1000

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] = 1; a[1] = 3; a[2] = 5; b[0] = 2; b[1] = 4; 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}]]];
u = Table[a[n], {n, 0, 100}]    (* A299490 *)
v = Table[b[n], {n, 0, 100}]    (* A299491 *)

Crossrefs

Cf. A022424, A299490.

Sequence in context: A039240 A039183 A039136 * A039098 A015860 A147613

Adjacent sequences: A299488 A299489 A299490 * A299492 A299493 A299494

Keyword

nonn,easy

Author

Clark Kimberling, Feb 16 2018

Status

approved