A296289 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + n*b(n-1), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.
1, 3, 12, 30, 66, 131, 245, 439, 764, 1302, 2196, 3652, 6028, 9888, 16154, 26312, 42770, 69422, 112570, 182410, 295440, 478354, 774344, 1253296, 2028288, 3282284, 5311326, 8594447, 13906669, 22502073, 36409762, 58912920, 95323834, 154237975, 249563101
Offset: 0
Examples
a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, b(2) = 5 a(2) = a(0) + a(1) + 2*b(1) = 12 Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, ...)
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
a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4; a[n_] := a[n] = a[n - 1] + a[n - 2] + n*b[n-1]; j = 1; While[j < 10, k = a[j] - j - 1; While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++]; Table[a[n], {n, 0, k}]; (* A296289 *) Table[b[n], {n, 0, 20}] (* complement *)
Comments