A294871 Solution of the complementary equation a(n) = 2*a(n-1) - a(n-2) + b(n-1) + 3, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.
1, 2, 10, 26, 51, 86, 132, 190, 262, 349, 452, 572, 710, 867, 1044, 1242, 1462, 1705, 1972, 2264, 2582, 2927, 3300, 3703, 4137, 4603, 5102, 5635, 6203, 6807, 7448, 8127, 8845, 9603, 10402, 11243, 12127, 13055, 14028, 15047, 16113, 17227, 18390, 19603, 20867
Offset: 0
Examples
a(0) = 1, a(1) = 2, b(0) = 3 b(1) = 4 (least "new number") a(2) = 2*a(1) - a(0) + b(1) + 3 = 10 Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, ...)
Links
- Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.
Crossrefs
Cf. A294860.
Programs
-
Mathematica
mex := First[Complement[Range[1, Max[#1] + 1], #1]] &; a[0] = 1; a[1] = 2; b[0] = 3; a[n_] := a[n] = 2 a[n - 1] - 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, 18}] (* A294871 *) Table[b[n], {n, 0, 10}]
Comments