A296223 Solution of the complementary equation a(n) = a(0)*b(n-1) + a(1)*b(n-2) + ... + a(n-1)*b(0) - 1, where a(0) = 1, a(1) = 3, b(0) = 2, and (a(n)) and (b(n)) are increasing complementary sequences.

1, 3, 9, 34, 124, 453, 1654, 6040, 22055, 80532, 294058, 1073735, 3920679, 14316124, 52274468, 190877084, 696976221, 2544966858, 9292793804, 33932079081, 123900951107, 452416889887, 1651973131976, 6032080786047, 22025781112962, 80425818360771

Offset: 0

Author

Clark Kimberling, Dec 10 2017

Keywords

• nonn
• easy

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295862 for a guide to related sequences.

Examples

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

Links

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

Crossrefs

Cf. A296000, A296224.

Programs

Mathematica

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}] - 1;
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
u = Table[a[n], {n, 0, 200}]  (* A296223 *)
Table[b[n], {n, 0, 20}]
N[Table[a[n]/a[n - 1], {n, 1, 200, 10}], 200];
RealDigits[Last[t], 10][[1]] (* A296224 *)