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A294477
Solution of the complementary equation a(n) = a(n-2) + b(n-1) + 2, where a(0) = 1, a(1) = 3, b(0) = 2.
2
1, 3, 7, 10, 15, 20, 26, 33, 40, 48, 56, 66, 75, 86, 96, 109, 120, 134, 146, 161, 175, 191, 206, 223, 239, 257, 275, 294, 313, 333, 353, 374, 396, 418, 441, 464, 488, 512, 537, 563, 589, 616, 643, 671, 699, 728, 758, 788, 819, 850, 882, 914, 947, 980, 1014
OFFSET
0,2
COMMENTS
The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A294476 for a guide to related sequences.
LINKS
Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.
EXAMPLE
a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
a(2) = a(0) + b(1) + 2 = 7
Complement: (b(n)) = (2, 4, 5, 6, 8, 9, 11, 12, 13, 14, 16,...)
MATHEMATICA
mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 3; b[0] = 2;
a[n_] := a[n] = a[n - 2] + b[n - 1] + 2;
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 40}] (* A294477 *)
Table[b[n], {n, 0, 10}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, Nov 01 2017
STATUS
approved