A245551 Expansion of 1/(1 - 2*x - 3*x^2)^(5/2).

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

Offset: 0

Author

N. J. A. Sloane, Jul 30 2014

Keywords

• nonn
• changed

Comments

From Petros Hadjicostas, Jun 03 2020: (Start)

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)

Examples

			From _Petros Hadjicostas_, Jun 03 2020: (Start)
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)
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Robert Coquereaux and Jean-Bernard Zuber, Counting partitions by genus. II. A compendium of results, arXiv:2305.01100 [math.CO], 2023. See p. 6.
• Lifoma Salaam, Combinatorial statistics on phylogenetic trees, Ph.D. Dissertation, Howard University, Washington D.C., 2008; see Theorem 39 (p. 25).
• J. Y. X. Yang, M. X. X. Zhong, and R. D. P. Zhou, On the Enumeration of (s, s+ 1, s+2)-Core Partitions, arXiv preprint arXiv:1406.2583 [math.CO], 2014. See Theorem 4.2.

Crossrefs

Cf. A000108, A001006, A002426, A102839, A374506.

Programs

Maple

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:
seq(A[n],n=0..100); # Robert Israel, Aug 01 2014

Mathematica

CoefficientList[Series[1/(1 - 2 x - 3 x^2)^(5/2), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 01 2014 *)

PARI

x='x+O('x^50); Vec(1/(1-2*x-3*x^2)^(5/2)) \\ G. C. Greubel, Apr 06 2017

Formula

a(n) ~ 3^(n+3/2) * n^(3/2) / (8*sqrt(Pi)). - Vaclav Kotesovec, Jul 31 2014

a(n) = (2+3/n)*a(n-1) + (3+9/n)*a(n-2) for n >= 2. - Robert Israel, Aug 01 2014

a(n) = (binomial(n+4,2)/6) * Sum_{k=0..floor(n/2)} binomial(n+2,n-2*k) * binomial(2*k+2,k). - Seiichi Manyama, Jul 10 2024

a(n) = Sum_{k=0..n} (-2)^k * (3/2)^(n-k) * binomial(-5/2,k) * binomial(k,n-k). - Seiichi Manyama, Aug 23 2025