A295756 Solution of the complementary equation a(n) = a(n-1) + a(n-3) + a(n-4) + b(n-2), 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, 14, 27, 43, 71, 123, 205, 332, 541, 885, 1439, 2330, 3775, 6119, 9909, 16036, 25953, 42005, 67975, 109990, 177976, 287985, 465980, 753977, 1219970, 1973968, 3193959, 5167941, 8361915, 13529879, 21891817, 35421712, 57313546, 92735283, 150048854

Offset: 0

Author

Clark Kimberling, Dec 01 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), so that a( ) has the growth rate of the Fibonacci numbers (A000045).

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) + a(1) + a(0) + b(2) = 14
Complement: (b(n)) = (5, 6, 7, 8, 9, 10, 11, 12, 13, 15, ...)
		

Links

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

Crossrefs

Cf. A001622, A000045, A293411, A295754.

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] + a[n - 3] + a[n - 4] + b[n - 2];
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}]   (* A295756 *)
Table[b[n], {n, 0, 20}]  (*complement *)