A008544 Triple factorial numbers: Product_{k=0..n-1} (3*k+2).

1, 2, 10, 80, 880, 12320, 209440, 4188800, 96342400, 2504902400, 72642169600, 2324549427200, 81359229952000, 3091650738176000, 126757680265216000, 5577337931669504000, 262134882788466688000

Offset: 0

Author

Joe Keane (jgk(AT)jgk.org)

Keywords

• nonn
• easy

Comments

a(n-1), n >= 1, enumerates increasing plane (aka ordered) trees with n vertices (one of them a root labeled 1) where each vertex with outdegree r >= 0 comes in r+1 types (like an (r+1)-ary vertex). See the increasing tree comments under A004747. - Wolfdieter Lang, Oct 12 2007

An example for the case of 3 vertices is shown below. For the enumeration of non-plane trees of this type see A029768. - Peter Bala, Aug 30 2011

a(n) is the product of the positive integers k <= 3*n that have k modulo 3 = 2. - Peter Luschny, Jun 23 2011

See A094638 for connections to differential operators. - Tom Copeland, Sep 20 2011

Partial products of A016789. - Reinhard Zumkeller, Sep 20 2013

The Mathar conjecture is true. Generally from the factorial form, the last term is the "extra" product beyond the prior term, from k=n-1 and 3k+2 evaluates to 3*(n-1)+2 = 3n-1, yielding a(n) = a(n-1)*(3n-1) (eqn1). Similarly, a(n) = a(n-2)*(3n-1)*(3(n-2)+2) = a(n-2)*(3n-1)*(3n-4) (eqn2) and a(n) = a(n-3)*(3n-1)*(3n-4)*(3*(n-2)+2) = a(n-3)*(3n-1)*(3n-4)*(3n-7) (eqn3). We equate (eqn2) and (eqn3) to get a(n-2)*(3n-1)*(3n-4) = a(n-3)*(3n-1)*(3n-4)*(3n-7) or a(n-2)+(7-3n)*a(n-3) = 0 (eqn4). From (eqn1) we have a(n)+(1-3n)*a(n-1) = 0 (eqn5). Combining (eqn4) and (eqn5) yields a(n)+(1-3n)*a(n-1)+a(n-2)+(7-3n)*a(n-3) = 0. - Bill McEachen, Jan 01 2016

a(n-1), n>=1, is the dimension of the n-th component of the operad encoding the multilinearization of the following identity in nonassociative algebras: s*(a,a,b)-(s+t)*(a,b,a)+t*(b,a,a)=0, for any given pair of scalars (s,t). Here (a,b,c) is the associator (ab)c-a(bc). This is proved in the referenced article on associator dependent algebras by Bremner and me. - Vladimir Dotsenko, Mar 22 2022

Examples

			a(2) = 10 from the described trees with 3 vertices: there are three trees with a root vertex (label 1) with outdegree r=2 (like the three 3-stars each with one different ray missing) and the four trees with a root (r=1 and label 1) a vertex with (r=1) and a leaf (r=0). Assigning labels 2 and 3 yields 2*3+4=10 such trees.
a(2) = 10. The 10 possible plane increasing trees on 3 vertices, where vertices of outdegree 1 come in 2 colors (denoted a or b) and vertices of outdegree 2 come in 3 colors (a, b or c), are:
.
   1a    1b    1a    1b        1a       1b       1c
   |     |     |     |        / \      / \      / \
   2a    2b    2b    2a      2   3    2   3    2   3
   |     |     |     |
   3     3     3     3         1a       1b       1c
                              / \      / \      / \
                             3   2    3   2    3   2
		

Links

• T. D. Noe, Table of n, a(n) for n = 0..100
• Murray Bremner and Vladimir Dotsenko, Associator dependent algebras and Koszul duality, arXiv:2203.11142 [math.RA], 2022.
• Wolfdieter Lang, On generalizations of Stirling number triangles, J. Integer Seq., Vol. 3 (2000), Article 00.2.4.
• Keiichi Shigechi, On the lattice of weighted partitions, arXiv:2212.14666 [math.CO], 2022. See p. 27.
• Garrett Southwood and Hua Wang, The Statistics and Combinatorics of Increasing Trees and Colored Increasing Trees, J. Combin. Math. Combin. Comput. (2024) Vol. 123, 475-487. See p. 485.

Crossrefs

a(n) = A004747(n+1, 1) (first column of triangle). Cf. A051141.

Cf. A000165, A001813, A047055, A047657, A084947, A084948, A084949.

Cf. A029768, A034724, A049308.

Cf. A024396, A193534.

Cf. A225470, A290596 (first columns).

Subsequence of A007661.

Programs

Haskell

a008544 n = a008544_list !! n
a008544_list = scanl (*) 1 a016789_list
-- Reinhard Zumkeller, Sep 20 2013

Magma

[Round((Gamma(2*n-5/3)/Gamma(n-5/6)*Gamma(2/3)/Gamma(5/6) )/ Sqrt(3)*3^n/4^(n-1)): n in [1..20]]; // Vincenzo Librandi, Feb 21 2015

Magma

[Round(3^n*Gamma(n+2/3)/Gamma(2/3)): n in [0..20]]; // G. C. Greubel, Mar 31 2019

Maple

a := n -> mul(3*k-1, k = 1..n);
A008544 := n -> mul(k, k = select(k-> k mod 3 = 2, [$1 .. 3*n])): seq(A008544(n), n = 0 .. 16); # Peter Luschny, Jun 23 2011

Mathematica

k = 3; b[1]=2; b[n_]:= b[n] = b[n-1]+k; a[0]=1; a[1]=2; a[n_]:= a[n] = a[n-1]*b[n]; Table[a[n], {n,0,20}] (* Roger L. Bagula, Sep 17 2008 *)
Product[3 k + 2, {k, 0, # - 1}] & /@ Range[0, 16] (* Michael De Vlieger, Jan 02 2016 *)
Table[3^n*Pochhammer[2/3, n], {n,0,20}] (* G. C. Greubel, Mar 31 2019 *)

Maxima

a(n):=((n)!*sum(binomial(k,n-k)*binomial(n+k,k)*3^(-n+k)*(-1)^(n-k),k,floor(n/2),n)); /* Vladimir Kruchinin, Sep 28 2013 */

PARI

a(n) = prod(k=0,n-1, 3*k+2 );

PARI

vector(20, n, n--; round(3^n*gamma(n+2/3)/gamma(2/3))) \\ G. C. Greubel, Mar 31 2019

Sage

@CachedFunction
def A008544(n): return 1 if n == 0 else (3*n-1)*A008544(n-1)
[A008544(n) for n in (0..16)]  # Peter Luschny, May 20 2013

Sage

[3^n*rising_factorial(2/3, n) for n in (0..20)] # G. C. Greubel, Mar 31 2019

Formula

a(n) = Product_{k=0..n-1} (3*k+2) = A007661(3*n-1) (with A007661(-1) = 1).

E.g.f.: (1-3*x)^(-2/3).

a(n) = 2*A034000(n), n >= 1, a(0) = 1.

a(n) ~ 2^(1/2)*Pi^(1/2)*Gamma(2/3)^-1*n^(1/6)*3^n*e^-n*n^n*{1 - 1/36*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Nov 22 2001

a(n) = (Gamma(2*n-5/3)/Gamma(n-5/6)*Gamma(2/3)/Gamma(5/6))/sqrt(3)*3^n/4^(n-1). - Jeremy L. Martin, Mar 31 2002 (typo fixed by Vincenzo Librandi, Feb 21 2015)

From Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003: (Start)

a(n) = A084939(n)/A000142(n)*A000079(n).

a(n) = 3^n*Pochhammer(2/3, n) = 3^n*Gamma(n+2/3)/Gamma(2/3). (End)

Let T = A094638 and c(t) = column vector(1, t, t^2, t^3, t^4, t^5,...), then A008544 = unsigned [ T * c(-3) ] and the list partition transform A133314 of [1,T * c(-3)] gives [1,T * c(3)] with all odd terms negated, which equals a signed version of A007559; i.e., LPT[(1,signed A008544)] = signed A007559. Also LPT[A007559] = (1,-A008544) and e.g.f. [1,T * c(t)] = (1-x*t)^(-1/t) for t = 3 or -3. Analogous results hold for the double factorial, quadruple factorial and so on. - Tom Copeland, Dec 22 2007

G.f.: 1/(1-2x/(1-3x/(1-5x/(1-6x/(1-8x/(1-9x/(1-11x/(1-12x/(1-...))))))))) (continued fraction). - Philippe Deléham, Jan 08 2012

a(n) = (-1)^n*Sum_{k=0..n} 3^k*s(n+1,n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012

G.f.: 1/Q(0) where Q(k) = 1 - x*(3*k+2)/(1 - x*(3*k+3)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 20 2013

G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(3*k+2)/(x*(3*k+2) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 25 2013

D-finite with recurrence: a(n) = (9*(n-2)*(n-1)+2)*a(n-2) + 4*a(n-1), n>=2. - Ivan N. Ianakiev, Aug 09 2013

a(n) = n!*Sum_{k=floor(n/2)..n} binomial(k,n-k)*binomial(n+k,k)*3^(-n+k)*(-1)^(n-k). - Vladimir Kruchinin, Sep 28 2013

Recurrence equation: a(n) = 3*a(n-1) + (3*n - 4)^2*a(n-2) with a(0) = 1 and a(1) = 2. A024396 satisfies the same recurrence (but with different initial conditions). This observation leads to a continued fraction expansion for the constant A193534 due to Euler. - Peter Bala, Feb 20 2015

a(n) = A225470(n, 0), n >= 0. - Wolfdieter Lang, May 29 2017

G.f.: Hypergeometric2F0(1, 2/3; -; 3*x). - G. C. Greubel, Mar 31 2019

D-finite with recurrence: a(n) + (-3*n+1)*a(n-1)=0. - R. J. Mathar, Jan 17 2020

G.f.: 1/(1-2*x-6*x^2/(1-8*x-30*x^2/(1-14*x-72*x^2/(1-20*x-132*x^2/(1-...))))) (Jacobi continued fraction). - Nikolaos Pantelidis, Feb 28 2020

G.f.: 1/G(0), where G(k) = 1 - (6*k+2)*x - 3*(k+1)*(3*k+2)*x^2/G(k+1). - Nikolaos Pantelidis, Feb 28 2020

Sum_{n>=0} 1/a(n) = 1 + (e/3)^(1/3) * (Gamma(2/3) - Gamma(2/3, 1/3)). - Amiram Eldar, Mar 01 2022