A299544 Solution b( ) of the complementary equation a(n) = 2*b(n-1) + b(n-2) - b(n-3), where a(0) = 1, a(1) = 2, a(2) = 3; see Comments.

4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 16, 18, 20, 22, 24, 26, 27, 28, 30, 31, 32, 33, 35, 36, 37, 39, 40, 41, 43, 44, 45, 47, 48, 49, 51, 52, 53, 55, 58, 59, 60, 62, 63, 66, 68, 69, 70, 72, 73, 76, 77, 78, 80, 81, 84, 85, 86, 88, 89, 92, 93, 94, 96, 97, 100, 101

Offset

0,1

Comments

From the Bode-Harborth-Kimberling link:

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

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] = 1; a[1] = 2; a[2] = 3; b[0] = 4; b[1] = 5;
a[n_] := 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}]]];
Table[a[n], {n, 0, 100}]    (* A299543 *)
Table[b[n], {n, 0, 100}]    (* A299544 *)

Crossrefs

Cf. A022424, A299543.

Sequence in context: A111222 A037357 A191842 * A039174 A016069 A360210

Adjacent sequences: A299541 A299542 A299543 * A299545 A299546 A299547

Keyword

nonn,easy

Author

Clark Kimberling, Feb 25 2018

Status

approved