A299545 Solution a( ) 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.

1, 2, 3, 12, 14, 16, 18, 20, 22, 25, 30, 34, 38, 42, 46, 49, 51, 55, 56, 58, 61, 65, 66, 69, 73, 74, 77, 81, 82, 85, 89, 90, 93, 97, 99, 104, 107, 108, 112, 119, 121, 123, 127, 128, 132, 138, 139, 143, 144, 148, 154, 155, 159, 160, 164, 170, 171, 175, 176

Offset: 0

Author

Clark Kimberling, Mar 01 2018

Keywords

• nonn
• easy

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

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

Crossrefs

Cf. A022424, A299546.

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] = 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 *)