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

0,1

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.

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 *)

Crossrefs

Cf. A022424 and references therein. Cf. A299540.

Sequence in context: A100036 A179781 A309370 * A193760 A113623 A308868

Adjacent sequences: A022435 A022436 A022437 * A022439 A022440 A022441

Keyword

nonn,easy

Author

Clark Kimberling

Status

approved