A299546 Solution b( ) of the complementary equation a(n) = b(n-1) + 2*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, 13, 15, 17, 19, 21, 23, 24, 26, 27, 28, 29, 31, 32, 33, 35, 36, 37, 39, 40, 41, 43, 44, 45, 47, 48, 50, 52, 53, 54, 57, 59, 60, 62, 63, 64, 67, 68, 70, 71, 72, 75, 76, 78, 79, 80, 83, 84, 86, 87, 88, 91, 92, 94, 95, 96, 98, 100, 101

Offset

0,1

Comments

From the Bode-Harborth-Kimberling link:

a(n) = b(n-1) + 2*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] = b[n - 1] + 2 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}]    (* A299545 *)
Table[b[n], {n, 0, 100}]    (* A299546 *)

Crossrefs

Cf. A022424, A299545.

Sequence in context: A016069 A360210 A194283 * A039128 A214421 A294237

Adjacent sequences: A299543 A299544 A299545 * A299547 A299548 A299549

Keyword

nonn,easy

Author

Clark Kimberling, Mar 01 2018

Status

approved