A296259 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2)^2, where a(0) = 2, a(1) = 3, b(0) = 1, and (a(n)) and (b(n)) are increasing complementary sequences.
2, 3, 6, 25, 56, 130, 250, 461, 811, 1393, 2348, 3910, 6454, 10589, 17299, 28177, 45800, 74338, 120538, 195317, 316339, 512185, 829100, 1341961, 2171790, 3514535, 5687166, 9202601, 14890728, 24094353, 38986170, 63081679, 102069074, 165152049, 267222492
Offset: 0
Examples
a(0) = 2, a(1) = 3, b(0) = 1; a(2) = a(0) + a(1) + b(0)^2 = 6; Complement: (b(n)) = (1, 4, 5, 7, 8, 9, 10, 11, 12, 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] = 2; a[1] = 3; b[0] = 1; a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n-2]^2; j = 1; While[j < 6 , 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}] (* A296259 *) Table[b[n], {n, 0, 20}] (* complement *)
Formula
a(n) = H + R, where H = f(n-1)*a(0) + f(n)*a(1) and R = f(n-1)*b(0)^2 + f(n-2)*b(1)^2 + ... + f(2)*b(n-3)^2 + f(1)*b(n-2)^2, where f(n) = A000045(n), the n-th Fibonacci number.
Comments