A295755 Solution of the complementary equation a(n) = a(n-1) + a(n-3) + a(n-4) + b(n-3), 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, 13, 25, 40, 66, 114, 190, 308, 502, 821, 1335, 2162, 3503, 5678, 9195, 14881, 24084, 38980, 63080, 102071, 165162, 267250, 432430, 699693, 1132136, 1831848, 2964004, 4795867, 7759886, 12555774, 20315682, 32871473, 53187172, 86058669, 139245866

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), so that a( ) has the growth rate of the Fibonacci numbers (A000045).

Links

Table of n, a(n) for n=0..37.

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

Example

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

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 - 3];
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}]   (* A295755 *)
Table[b[n], {n, 0, 20}]  (*complement *)

Crossrefs

Cf. A001622, A000045, A293411, A295754.

Sequence in context: A162222 A010346 A295621 * A089142 A123215 A051855

Adjacent sequences: A295752 A295753 A295754 * A295756 A295757 A295758

Keyword

nonn,easy

Author

Clark Kimberling, Nov 30 2017

Status

approved