A296254 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1)^2, where a(0) = 2, a(1) = 3, b(0) = 1, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.
2, 3, 21, 49, 106, 204, 374, 659, 1133, 1913, 3190, 5272, 8658, 14155, 23069, 37513, 60906, 98780, 160086, 259350, 419965, 679891, 1100481, 1781048, 2882258, 4664090, 7547189, 12212179, 19760329, 31973532, 51734950, 83709638, 135445813, 219156747, 354603929
Offset: 0
Examples
a(0) = 2, a(1) = 3, b(0) = 1, b(1) = 4; a(2) = a(0) + a(1) + b(1)^2 = 21; Complement: (b(n)) = (1, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, ...)
Links
- Clark Kimberling, Table of n, a(n) for n = 0..1000
- Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.
Programs
-
Mathematica
a[0] = 2; a[1] = 3; b[0] = 1; b[1] = 4; a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n-1]^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}] (* A296254 *) 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(1)^2 + f(n-2)*b(2)^2 + ... + f(2)*b(n-2)^2 + f(1)*b(n-1)^2, where f(n) = A000045(n), the n-th Fibonacci number.
Comments