A296252 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1)^2, 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, 20, 48, 104, 201, 369, 651, 1120, 1892, 3156, 5217, 8569, 14011, 22836, 37136, 60296, 97793, 158530, 256807, 415866, 673249, 1089740, 1763665, 2854134, 4618583, 7473558, 12093041, 19567560, 31661625, 51230274, 82893055, 134124554, 217018905, 351144828

Offset: 0

Author

Clark Kimberling, Dec 11 2017

Keywords

• nonn
• easy

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622). See A296245 for a guide to related sequences.

Examples

			a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4;
a(2) = a(0) + a(1) + b(1)^2 = 20;
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, ...)
		

Links

• Clark Kimberling, Table of n, a(n) for n = 0..1000
• Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.

Crossrefs

Cf. A001622, A296245.

Programs

Mathematica

a[0] = 1; a[1] = 3; b[0] = 2; 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}]   (* A296252 *)
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.