A295360 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - 3*b(n-3), where a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.

1, 3, 5, 12, 18, 27, 41, 63, 98, 155, 247, 392, 628, 1008, 1624, 2620, 4228, 6831, 11041, 17853, 28874, 46706, 75559, 122244, 197778, 319996, 517747, 837715, 1355433, 2193118, 3548520, 5741606, 9290093, 15031665, 24321723

Offset: 0

Author

Clark Kimberling, Nov 21 2017

Keywords

• nonn
• easy

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295357 for a guide to related sequences.

Examples

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

Links

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

Crossrefs

Cf. A001622, A295357.

Programs

Mathematica

mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 3; a[2] = 5; b[0] = 2; b[1] = 4; b[2] = 6;
a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + b[n - 2] - 3*b[n - 3];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
z = 32; u = Table[a[n], {n, 0, z}]   (* A295360 *)
v = Table[b[n], {n, 0, 10}]  (* complement *)

Formula

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).