A022438 a(n) = c(n-1) + c(n-3) where c is the sequence of numbers not in a.

2, 3, 5, 7, 12, 15, 18, 20, 23, 25, 29, 31, 35, 38, 41, 45, 48, 51, 54, 57, 60, 63, 66, 69, 71, 75, 77, 81, 83, 86, 89, 91, 95, 97, 101, 103, 107, 109, 113, 115, 119, 121, 125, 127, 131, 133, 137, 140, 143, 146, 149, 152, 155, 158, 161, 164, 167, 171, 173

Offset: 0

Author

Clark Kimberling

Keywords

• nonn
• easy

Comments

From Clark Kimberling, Feb 25 2018: (Start)

Solution a( ) of the complementary equation a(n) = b(n-1) + b(n-3), where a(0) = 2, a(1) = 3, a(2) = 5.

From the Bode-Harborth-Kimberling link:

a(n) = b(n-1) + 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.

(End)

Links

• Ivan Neretin, Table of n, a(n) for n = 0..10000
• J-P. Bode, H. Harborth, C. Kimberling, Complementary Fibonacci sequences, Fibonacci Quarterly 45 (2007), 254-264.

Crossrefs

Cf. A022424 and references therein. Cf. A299540.

Programs

Mathematica

Fold[Append[#1, Plus @@ Complement[Range[Max@#1 + 3], #1][[{#2, #2 + 2}]]] &, {2, 3, 5}, Range[43]] (* Ivan Neretin, Mar 30 2017 *)
mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 2; a[1] = 3; a[2] = 5; b[0] = 1; b[1] = 4;
a[n_] := a[n] = b[n - 1] + 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}]    (* A022438 *)
Table[b[n], {n, 0, 100}]    (* A299540 *)
(* Clark Kimberling, Feb 25 2018 *)