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A296000
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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.
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23
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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
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OFFSET
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0,2
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COMMENTS
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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
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LINKS
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EXAMPLE
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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, ...)
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MATHEMATICA
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$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 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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