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

0,2

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.

Links

Clark Kimberling, Table of n, a(n) for n = 0..1000

Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.

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.

Example

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, ...)

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 *)

Crossrefs

Cf. A001622, A296245.

Sequence in context: A344890 A072472 A267055 * A211068 A343010 A360417

Adjacent sequences: A296249 A296250 A296251 * A296253 A296254 A296255

Keyword

nonn,easy

Author

Clark Kimberling, Dec 11 2017

Status

approved