A296003 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) = 2, a(1) = 4, b(0) = 1, and (a(n)) and (b(n)) are increasing complementary sequences.

2, 4, 10, 32, 94, 278, 824, 2440, 7228, 21408, 63406, 187800, 556234, 1647478, 4879574, 14452538, 42806168, 126785206, 375518042, 1112225982, 3294240212, 9757026674, 28898794076, 85593729210, 253515301048, 750872855508, 2223968505284, 6587048494582

Offset: 0

Author

Clark Kimberling, Dec 07 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) -> 2.961844324... (as in A296004). See A296000 for a guide to related sequences.

Examples

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

Links

• Clark Kimberling, Table of n, a(n) for n = 0..999
• Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.

Crossrefs

Cf. A296000, A296004.

Programs

Mathematica

mex[list_] := NestWhile[# + 1 &, 1, MemberQ[list, #] &];
a[0] = 2; a[1] = 4; b[0] = 1; 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}];  (* A296003 *)
t = N[Table[a[n]/a[n - 1], {n, 1, 500, 100}], 200]
Take[RealDigits[Last[t], 10][[1]], 100]  (* A296004 *)