A296005 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) = 3, b(0) = 1, and (a(n)) and (b(n)) are increasing complementary sequences.

2, 3, 11, 33, 104, 323, 1007, 3136, 9769, 30431, 94791, 295274, 919773, 2865082, 8924690, 27800290, 86597525, 269750118, 840267961, 2617423311, 8153238141, 25397226311, 79112015761, 246432856920, 767635009499, 2391172651130, 7448470401642, 23201884354901

Offset

0,1

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. a(n)/a(n-1) -> 3.114986447390302... (as in A296006). See A296000 for a guide to related sequences.

Links

Table of n, a(n) for n=0..27.

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

Example

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

Mathematica

mex[list_] := NestWhile[# + 1 &, 1, MemberQ[list, #] &];
a[0] = 2; a[1] = 3; 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}];  (* A296005 *)
t = N[Table[a[n]/a[n - 1], {n, 1, 500, 100}], 200]
Take[RealDigits[Last[t], 10][[1]], 100]  (* A296006 *)

Crossrefs

Cf. A296000, A296006.

Sequence in context: A268687 A062630 A300771 * A374310 A159458 A305846

Adjacent sequences: A296002 A296003 A296004 * A296006 A296007 A296008

Keyword

nonn,easy

Author

Clark Kimberling, Dec 07 2017

Extensions

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

Status

approved