A296001 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) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.

1, 2, 10, 43, 185, 796, 3425, 14737, 63411, 272845, 1174000, 5051498, 21735632, 93524277, 402417118, 1731524071, 7450417675, 32057725596, 137938276110, 593522081260, 2553812262104, 10988566855385, 47281706383454, 203444160458068, 875381402033582

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) -> 4.302809183918588... (as in A296002). See A296000 for a guide to related sequences.

Examples

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

Links

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

Crossrefs

Cf. A296000, A296002.

Programs

Mathematica

mex[list_] := NestWhile[# + 1 &, 1, MemberQ[list, #] &];
a[0] = 1; a[1] = 2; b[0] = 3; 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}];  (* A296001 *)
t = N[Table[a[n]/a[n - 1], {n, 1, 500, 100}], 200]
Take[RealDigits[Last[t], 10][[1]], 100]  (* A296002 *)

Extensions

Conjectured g.f. removed by Alois P. Heinz, Jun 25 2018