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A245551
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G.f.: 1/(1-2*x-3*x^2)^(5/2).
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1
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1, 5, 25, 105, 420, 1596, 5880, 21120, 74415, 258115, 883883, 2994355, 10051860, 33479460, 110750580, 364177332, 1191186855, 3877914915, 12571302975, 40598200335, 130657125984, 419173385400, 1340928798300, 4278305877300, 13617034683525, 43243221276801, 137040737988105
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OFFSET
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0,2
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COMMENTS
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For n >= 4, 2*a(n-4) counts 3-sets of leaves in "0,1,2" Motzkin rooted trees with n edges. "0,1,2" trees are rooted trees where each vertex has out-degree zero, one, or two. They are counted by the Motzkin numbers A001006.
For "0,1,2" trees, Salaam (2008) proved that the g.f. of the number of r-sets of leaves is A000108(r-1) * z^(2*r-2) * T(z)^(2*r-1), where T(z) = 1/sqrt(1 - 2*z - 3*z^2) is the g.f. of the central trinomial numbers A002426.
For r = 2, we get a shifted version of A102839. For r = 3, we get twice of a shifted version of the current sequence. (End)
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LINKS
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FORMULA
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a(n) = (2+3/n)*a(n-1) + (3+9/n)*a(n-2) for n >= 2. - Robert Israel, Aug 01 2014
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EXAMPLE
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Out of the A001006(4) = 9 Motzkin trees with n = 4 edges, only the following 2*a(4-4) = 2 have 3-sets of leaves:
A A
/ \ / \
/ \ / \
B C B C
/ \ / \
/ \ / \
D E D E
{C, D, E} {B, D, E}
(End)
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MAPLE
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A[0]:= 1: A[1]:= 5:
for n from 2 to 100 do
A[n]:= (2+3/n)*A[n-1] + (3+9/n)*A[n-2]
od:
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MATHEMATICA
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CoefficientList[Series[1/(1 - 2 x - 3 x^2)^(5/2), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 01 2014 *)
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PROG
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(PARI) x='x+O('x^50); Vec(1/(1-2*x-3*x^2)^(5/2)) \\ G. C. Greubel, Apr 06 2017
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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