A294476 Solution of the complementary equation a(n) = a(n-2) + b(n-1) + 1, where a(0) = 1, a(1) = 3, b(0) = 2.
1, 3, 6, 9, 14, 18, 25, 30, 38, 44, 54, 61, 72, 81, 93, 103, 116, 127, 141, 154, 169, 183, 199, 215, 232, 249, 267, 285, 304, 323, 344, 364, 386, 407, 430, 453, 477, 501, 526, 551, 577, 603, 630, 657, 686, 714, 744, 773, 804, 834, 867, 898, 932, 964, 999
Offset: 0
Examples
a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that a(2) = a(0) + b(1) + 1 = 6 Complement: (b(n)) = (2, 4, 5, 7, 8, 10, 11, 12, 13, 15,...)
Links
- Clark Kimberling, Table of n, a(n) for n = 0..1000
- Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.
Programs
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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] + 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}] (* A294476 *) Table[b[n], {n, 0, 10}]
Comments