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A296000 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) = 3, b(0) = 2, and (a(n)) and (b(n)) are increasing complementary sequences. 23
1, 3, 10, 37, 135, 493, 1800, 6572, 23996, 87614, 319895, 1167997, 4264577, 15570774, 56851829, 207576737, 757901769, 2767242128, 10103722287, 36890593353, 134694505577, 491795012865, 1795636233585, 6556206140806, 23937943641806, 87401941533192 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. a(n)/a(n-1) -> 3.651188... (as in A295999). Guide for the complementary equation a(n) = a(0)*b(n-1) + a(1)*b(n-2) + ... + a(n-1)*b(0):
A296000: a(0) = 1, a(1) = 3, b(0) = 2, limiting ratio of a(n)/a(n-1): A295999
A296001: a(0) = 1, a(1) = 2, b(0) = 3, limiting ratio of a(n)/a(n-1): A296002
A296003: a(0) = 2, a(1) = 4, b(0) = 1, limiting ratio of a(n)/a(n-1): A296004
A296005: a(0) = 2, a(1) = 3, b(0) = 1, limiting ratio of a(n)/a(n-1): A296006
LINKS
Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.
EXAMPLE
a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
a(2) = a(0)*b(1) + a(1)*b(0) = 10
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, ...)
MATHEMATICA
$RecursionLimit = Infinity;
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}];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 100}] (* A296000 *)
t = N[Table[a[n]/a[n - 1], {n, 1, 500, 100}], 200]
Take[RealDigits[Last[t], 10][[1]], 100] (* A295999 *)
CROSSREFS
Sequence in context: A359721 A138807 A149043 * A360586 A242725 A151315
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Dec 04 2017
EXTENSIONS
Incorrect conjectured g.f. removed by Georg Fischer, Sep 23 2020
STATUS
approved

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Last modified March 29 02:23 EDT 2024. Contains 371264 sequences. (Running on oeis4.)