A294551 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + 1, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.
1, 2, 11, 23, 46, 83, 145, 246, 411, 680, 1117, 1825, 2972, 4829, 7835, 12700, 20573, 33313, 53928, 87285, 141260, 228595, 369907, 598556, 968519, 1567133, 2535712, 4102907, 6638683, 10741656, 17380407, 28122133, 45502612, 73624819, 119127507, 192752404
Offset: 0
Examples
a(0) = 1, a(1) = 2, b(0) = 3, so that b(1) = 4 (least "new number"); a(2) = a(1) + a(0) + b(1) + b(0) + 1 = 11. Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, ...).
Links
- 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 - 1] + a[n - 2] + b[n - 1] + b[n - 2] + 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}] (* A294551 *) Table[b[n], {n, 0, 10}]
Comments