A294562 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, 5, 10, 17, 29, 48, 80, 130, 212, 344, 558, 904, 1465, 2371, 3838, 6211, 10051, 16264, 26317, 42583, 68902, 111487, 180391, 291881, 472274, 764157, 1236433, 2000592, 3237027, 5237621, 8474650, 13712273, 22186925, 35899200, 58086127
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 = 5 Complement: (b(n)) = (3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 18, ...)
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}] (* A294562 *) Table[b[n], {n, 0, 10}]
Comments