A296000 Solution of the complementary equation a(n) = a(0)*b(n-1) + a(1)*b(n-2) + ... + a(n-1)*b(0), where a(0) = 1, a(1) = 3, b(0) = 2, and (a(n)) and (b(n)) are increasing complementary sequences.

1, 3, 10, 37, 135, 493, 1800, 6572, 23996, 87614, 319895, 1167997, 4264577, 15570774, 56851829, 207576737, 757901769, 2767242128, 10103722287, 36890593353, 134694505577, 491795012865, 1795636233585, 6556206140806, 23937943641806, 87401941533192

Offset: 0

Author

Clark Kimberling, Dec 04 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) -> 3.651188... (as in A295999). Guide for the complementary equation a(n) = a(0)*b(n-1) + a(1)*b(n-2) + ... + a(n-1)*b(0):

A296000: a(0) = 1, a(1) = 3, b(0) = 2, limiting ratio of a(n)/a(n-1): A295999

A296001: a(0) = 1, a(1) = 2, b(0) = 3, limiting ratio of a(n)/a(n-1): A296002

A296003: a(0) = 2, a(1) = 4, b(0) = 1, limiting ratio of a(n)/a(n-1): A296004

A296005: a(0) = 2, a(1) = 3, b(0) = 1, limiting ratio of a(n)/a(n-1): A296006

Examples

			a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
a(2) = a(0)*b(1) + a(1)*b(0) = 10
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, ...)
		

Links

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

Crossrefs

Cf. A295999, A296001.

Programs

Mathematica

$RecursionLimit = Infinity;
mex[list_] := NestWhile[# + 1 &, 1, MemberQ[list, #] &];
a[0] = 1; a[1] = 3; b[0] = 2; a[n_] := a[n] = Sum[a[k]*b[n - k - 1], {k, 0, 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, 100}]  (* A296000 *)
t = N[Table[a[n]/a[n - 1], {n, 1, 500, 100}], 200]
Take[RealDigits[Last[t], 10][[1]], 100]  (* A295999 *)

Extensions

Incorrect conjectured g.f. removed by Georg Fischer, Sep 23 2020