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

0,2

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(2) = 14
Complement: (b(n)) = (5, 6, 7, 8, 9, 10, 11, 12, 13, 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 - 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 *)

Crossrefs

Cf. A001622, A000045, A293411, A295754.

Sequence in context: A188533 A163128 A280924 * A140047 A297840 A102483

Adjacent sequences: A295753 A295754 A295755 * A295757 A295758 A295759

Keyword

nonn,easy

Author

Clark Kimberling, Dec 01 2017

Status

approved