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A299494
Solution a( ) of the complementary equation a(n) = b(n-1) + b(n-2) + b(n-3), where a(0) = 2, a(1) = 4, a(2) = 6; see Comments.
3
2, 4, 6, 9, 15, 20, 25, 29, 33, 36, 39, 43, 47, 51, 54, 58, 62, 66, 69, 73, 77, 81, 85, 89, 93, 97, 101, 106, 110, 115, 119, 123, 127, 131, 135, 139, 143, 147, 151, 155, 160, 164, 168, 172, 176, 180, 184, 188, 192, 196, 200, 205, 209, 213, 217, 221, 225, 229
OFFSET
0,1
COMMENTS
From the Bode-Harborth-Kimberling link:
a(n) = b(n-1) + b(n-2) + b(n-3) for n > 2;
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.
MATHEMATICA
mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 2; a[1] = 4; a[2] = 6; b[0] = 1; b[1] = 3; b[2] = 5;
a[n_] := a[n] = b[n - 1] + 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}] (* A299494 *)
Table[b[n], {n, 0, 100}] (* A299495 *)
CROSSREFS
Sequence in context: A323227 A283024 A090483 * A306597 A127740 A323432
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Feb 21 2018
STATUS
approved