A299543 Solution a( ) 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.

1, 2, 3, 13, 15, 17, 19, 21, 23, 25, 29, 34, 38, 42, 46, 50, 54, 56, 57, 61, 64, 65, 67, 71, 74, 75, 79, 82, 83, 87, 90, 91, 95, 98, 99, 103, 106, 107, 111, 118, 121, 121, 125, 128, 133, 139, 140, 141, 145, 148, 153, 157, 157, 161, 164, 169, 173, 173, 177

Offset: 0

Author

Clark Kimberling, Feb 25 2018

Keywords

• nonn
• easy

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

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

Crossrefs

Cf. A022424, A299544.

Programs

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 *)