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A294482
Solution of the complementary equation a(n) = a(n-2) + b(n-1) + n + 1, where a(0) = 1, a(1) = 3, b(0) = 2.
2
1, 3, 8, 12, 19, 25, 35, 43, 55, 66, 80, 93, 109, 124, 142, 160, 180, 200, 222, 244, 269, 293, 320, 346, 375, 403, 434, 464, 497, 530, 565, 600, 637, 674, 713, 752, 794, 835, 879, 922, 968, 1013, 1061, 1108, 1158, 1207, 1259, 1311, 1365, 1419, 1475, 1531
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, 11, 13, 15, ...)
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] + n + 1;
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 40}] (* A294482 *)
Table[b[n], {n, 0, 10}]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Nov 03 2017
STATUS
approved