A295063 Solution of the complementary equation a(n) = 4*a(n-2) + b(n-1) + b(n-2), where a(0) = 1, a(1) = 3, b(0) = 2, and (a(n)) and (b(n)) are increasing complementary sequences.
1, 3, 10, 21, 51, 97, 219, 405, 896, 1643, 3609, 6599, 14465, 26427, 57893, 105743, 231609, 423011, 926478, 1692089, 3705959, 6768405, 14823887, 27073673, 59295603, 108294749, 237182471
Offset: 0
Examples
a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4 a(2) = 4*a(0) + b(1) + b(0) = 10 Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, ...)
Links
- Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.
Crossrefs
Cf. A295053.
Programs
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Mathematica
mex := First[Complement[Range[1, Max[#1] + 1], #1]] &; a[0] = 1; a[1] = 3; b[0] = 2; b[1]=4; a[n_] := a[n] = 4 a[n - 2] + b[n - 1] + b[n - 2]; b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]]; Table[a[n], {n, 0, 18}] (* A295063 *) Table[b[n], {n, 0, 10}]
Comments