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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.
3
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
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
Sequence in context: A268687 A062630 A300771 * A374310 A159458 A305846
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Dec 07 2017
EXTENSIONS
Conjectured g.f. removed by Alois P. Heinz, Jun 25 2018
STATUS
approved