A293358 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

1, 3, 8, 16, 30, 53, 92, 155, 258, 425, 696, 1135, 1846, 2998, 4862, 7879, 12761, 20661, 33444, 54128, 87596, 141749, 229371, 371147, 600546, 971722, 1572299, 2544053, 4116385, 6660472, 10776892, 17437400, 28214329, 45651767, 73866135, 119517942

Offset

0,2

Comments

The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:

A293358: a(n) = a(n-1) + a(n-2) + b(n-1)

A293406: a(n) = a(n-1) + a(n-2) + b(n-1) + 1

A293765: a(n) = a(n-1) + a(n-2) + b(n-1) + 2

A293766: a(n) = a(n-1) + a(n-2) + b(n-1) + 3

A293767: a(n) = a(n-1) + a(n-2) + b(n-1) - 1

A294365: a(n) = a(n-1) + a(n-2) + b(n-1) + n

A294366: a(n) = a(n-1) + a(n-2) + b(n-1) + 2n

A294367: a(n) = a(n-1) + a(n-2) + b(n-1) + n - 1

A294368: a(n) = a(n-1) + a(n-2) + b(n-1) + n + 1

Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio.

Links

Table of n, a(n) for n=0..35.

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

Example

a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
a(2) = a(1) + a(0) + b(1) = 8;
Complement: (b(n)) = (2, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, ...)

Mathematica

mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 40}]  (* A293358 *)
Table[b[n], {n, 0, 10}]

Crossrefs

Cf. A001622 (golden ratio), A293076.

Sequence in context: A009439 A000233 A002624 * A227265 A295960 A068039

Adjacent sequences: A293355 A293356 A293357 * A293359 A293360 A293361

Keyword

nonn,easy

Author

Clark Kimberling, Oct 29 2017

Status

approved