A295948 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n), where a(0) = 3, a(1) = 4, b(0) = 1, b(1) = 2, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.

3, 4, 12, 22, 41, 71, 121, 202, 334, 549, 897, 1461, 2374, 3852, 6244, 10115, 16379, 26515, 42917, 69456, 112398, 181880, 294305, 476213, 770547, 1246790, 2017368, 3264190, 5281591, 8545815, 13827441, 22373292, 36200770, 58574100, 94774909, 153349049

Offset: 0

Author

Clark Kimberling, Dec 08 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 A295862 for a guide to related sequences.

Examples

			a(0) = 3, a(1) = 4, b(0) = 1, b(1) = 2, b(2) = 5, so that
b(3) = 6 (least "new number");
a(2) = a(1) + a(0) + b(2) = 12;
Complement: (b(n)) = (1, 2, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, ...)
		

Links

• Clark Kimberling, Table of n, a(n) for n = 0..2000
• Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.

Crossrefs

Cf. A001622, A000045, A295862.

Programs

Mathematica

a[0] = 3; a[1] = 4; b[0] = 1; b[1] = 2; b[2] = 5;
a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n];
j = 1; While[j < 16, 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}]  (* A295948 *)
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(2) + f(n-2)*b(3) + ... + f(2)*b(n-1) + f(1)*b(n), where f(n) = A000045(n), the n-th Fibonacci number.