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A294478
Solution of the complementary equation a(n) = a(n-2) + b(n-1) + 3, where a(0) = 1, a(1) = 3, b(0) = 2.
2
1, 3, 8, 11, 17, 21, 29, 34, 44, 50, 61, 68, 80, 89, 102, 112, 127, 138, 154, 166, 183, 196, 214, 229, 248, 264, 284, 302, 323, 342, 364, 384, 407, 428, 452, 474, 500, 523, 550, 574, 602, 628, 657, 684, 714, 742, 773, 802, 834, 864, 897, 929, 963, 996, 1031
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) + 3 = 8
Complement: (b(n)) = (2, 4, 5, 6, 7, 9, 10, 12, 13, 14, ...)
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] + 3;
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 40}] (* A294478 *)
Table[b[n], {n, 0, 10}]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Nov 01 2017
STATUS
approved