A295620 Solution of the complementary equation a(n) = a(n-1) + 3*a(n-2) -2*a(n-3) - 2*a(n-4) + b(n-4), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, and (a(n)) and (b(n)) are increasing complementary sequences.

1, 2, 3, 4, 12, 20, 49, 85, 177, 304, 578, 979, 1765, 2953, 5150, 8538, 14570, 23997, 40352, 66149, 110094, 179867, 297172, 484313, 795934, 1294823, 2119684, 3443689, 5621258, 9123343, 14860404, 24100573, 39192618, 63526879, 103182816, 167177109, 271286602

Offset: 0

Author

Clark Kimberling, Nov 25 2017

Keywords

• nonn
• easy

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values.

Examples

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

Links

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

Crossrefs

Cf. A001622, A000045, A295619, A295621.

Programs

Mathematica

mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 2; a[2] = 3; a[3] = 4;
b[0] = 5; b[1] = 6; b[2] = 7; b[3] = 8;
a[n_] := a[n] = a[n - 1] + 3*a[n - 2] - 2*a[n - 3] - 2 a[n - 4] + b[n - 4];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
z = 36;  Table[a[n], {n, 0, z}]   (* A295620 *)
Table[b[n], {n, 0, 20}]  (*complement *)