A296258 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2)^2, where a(0) = 1, a(1) = 3, b(0) = 2, and (a(n)) and (b(n)) are increasing complementary sequences.
1, 3, 8, 27, 60, 123, 232, 436, 768, 1325, 2237, 3731, 6164, 10120, 16540, 26949, 43813, 71123, 115336, 186900, 302720, 490149, 793445, 1284219, 2078340, 3363343, 5442524, 8806767, 14250252, 23058043, 37309384, 60368583, 97679192, 158049071, 255729632
Offset: 0
Examples
a(0) = 1, a(1) = 3, b(0) = 2; a(2) = a(0) + a(1) + b(0)^2 = 8; Complement: (b(n)) = (2, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, ...)
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] = 1; a[1] = 3; b[0] = 2; 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}] (* A296258 *) 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