A295757 Solution of the complementary equation a(n) = a(n-1) + a(n-3) + a(n-4) + b(n-1), 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, 15, 29, 46, 76, 132, 220, 356, 580, 949, 1543, 2498, 4047, 6560, 10623, 17191, 27822, 45030, 72870, 117910, 190790, 308720, 499531, 808263, 1307806, 2116091, 3423920, 5540025, 8963959, 14504008, 23467992, 37972016, 61440024, 99412066, 160852117

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(3) = 15
Complement: (b(n)) = (5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, ...)

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 - 1];
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}]   (* A295757 *)
Table[b[n], {n, 0, 20}]  (*complement *)

Crossrefs

Cf. A001622, A000045, A293411, A295754.

Sequence in context: A269725 A270929 A103096 * A005645 A362639 A348628

Adjacent sequences: A295754 A295755 A295756 * A295758 A295759 A295760

Keyword

nonn,easy

Author

Clark Kimberling, Dec 01 2017

Status

approved