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A022438 a(n) = c(n-1) + c(n-3) where c is the sequence of numbers not in a. 3
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 (list; graph; refs; listen; history; text; internal format)
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
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
KEYWORD
nonn,easy
AUTHOR
STATUS
approved

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Last modified May 23 04:41 EDT 2024. Contains 372758 sequences. (Running on oeis4.)