A299416 Solution a( ) of the complementary equation a(n) = b(n-1) + b(n-2), where a(0) = 3, a(1) = 4; see Comments.

3, 4, 3, 7, 11, 14, 17, 19, 22, 25, 28, 31, 34, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 73, 76, 79, 82, 85, 88, 91, 94, 97, 100, 103, 106, 109, 112, 115, 118, 121, 124, 127, 130, 133, 136, 139, 142, 146, 149, 152, 155, 158, 161, 164, 167, 170, 173

Offset: 0

Author

Clark Kimberling, Feb 15 2018

Keywords

• nonn
• easy

Comments

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

b(0) = least positive integer not in {a(0),a(1)};

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..2000
• J-P. Bode, H. Harborth, C. Kimberling, Complementary Fibonacci sequences, Fibonacci Quarterly 45 (2007), 254-264.

Crossrefs

Cf. A022424, A299417.

Programs

Mathematica

mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 3; a[1] = 4; b[0] = 1; b[1] = 2;
a[n_] := a[n] = b[n - 1] + b[n - 2];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 100}]  (* A299416 *)
Table[b[n], {n, 0, 100}]  (* A299417 *)