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%I #163 Jan 02 2023 05:07:28
%S 1,2,10,80,880,12320,209440,4188800,96342400,2504902400,72642169600,
%T 2324549427200,81359229952000,3091650738176000,126757680265216000,
%U 5577337931669504000,262134882788466688000
%N Triple factorial numbers: Product_{k=0..n-1} (3*k+2).
%C 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
%C 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
%C a(n) is the product of the positive integers k <= 3*n that have k modulo 3 = 2. - _Peter Luschny_, Jun 23 2011
%C See A094638 for connections to differential operators. - _Tom Copeland_, Sep 20 2011
%C Partial products of A016789. - _Reinhard Zumkeller_, Sep 20 2013
%C 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
%C 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
%H T. D. Noe, <a href="/A008544/b008544.txt">Table of n, a(n) for n = 0..100</a>
%H Murray Bremner and Vladimir Dotsenko, <a href="https://arxiv.org/abs/2203.11142">Associator dependent algebras and Koszul duality</a>, arXiv:2203.11142 [math.RA], 2022.
%H Wolfdieter Lang, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL3/LANG/lang.html">On generalizations of Stirling number triangles</a>, J. Integer Seq., Vol. 3 (2000), Article 00.2.4.
%H Keiichi Shigechi, <a href="https://arxiv.org/abs/2212.14666">On the lattice of weighted partitions</a>, arXiv:2212.14666 [math.CO], 2022. See p. 27.
%F a(n) = Product_{k=0..n-1} (3*k+2) = A007661(3*n-1) (with A007661(-1) = 1).
%F E.g.f.: (1-3*x)^(-2/3).
%F a(n) = 2*A034000(n), n >= 1, a(0) = 1.
%F 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
%F 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)
%F From Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003: (Start)
%F a(n) = A084939(n)/A000142(n)*A000079(n).
%F a(n) = 3^n*Pochhammer(2/3, n) = 3^n*Gamma(n+2/3)/Gamma(2/3). (End)
%F 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
%F 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
%F 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
%F 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
%F 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
%F 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
%F 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
%F 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
%F a(n) = A225470(n, 0), n >= 0. - _Wolfdieter Lang_, May 29 2017
%F G.f.: Hypergeometric2F0(1, 2/3; -; 3*x). - _G. C. Greubel_, Mar 31 2019
%F D-finite with recurrence: a(n) + (-3*n+1)*a(n-1)=0. - _R. J. Mathar_, Jan 17 2020
%F 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
%F 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
%F Sum_{n>=0) 1/a(n) = 1 + (e/3)^(1/3) * (Gamma(2/3) - Gamma(2/3, 1/3)). - _Amiram Eldar_, Mar 01 2022
%e 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.
%e 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:
%e .
%e 1a 1b 1a 1b 1a 1b 1c
%e | | | | / \ / \ / \
%e 2a 2b 2b 2a 2 3 2 3 2 3
%e | | | |
%e 3 3 3 3 1a 1b 1c
%e / \ / \ / \
%e 3 2 3 2 3 2
%p a := n -> mul(3*k-1, k = 1..n);
%p 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
%t 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 *)
%t Product[3 k + 2, {k, 0, # - 1}] & /@ Range[0, 16] (* _Michael De Vlieger_, Jan 02 2016 *)
%t Table[3^n*Pochhammer[2/3, n], {n,0,20}] (* _G. C. Greubel_, Mar 31 2019 *)
%o (PARI) a(n) = prod(k=0,n-1, 3*k+2 );
%o (PARI) vector(20, n, n--; round(3^n*gamma(n+2/3)/gamma(2/3))) \\ _G. C. Greubel_, Mar 31 2019
%o (Sage)
%o @CachedFunction
%o def A008544(n): return 1 if n == 0 else (3*n-1)*A008544(n-1)
%o [A008544(n) for n in (0..16)] # _Peter Luschny_, May 20 2013
%o (Sage) [3^n*rising_factorial(2/3, n) for n in (0..20)] # _G. C. Greubel_, Mar 31 2019
%o (Haskell)
%o a008544 n = a008544_list !! n
%o a008544_list = scanl (*) 1 a016789_list
%o -- _Reinhard Zumkeller_, Sep 20 2013
%o (Maxima)
%o 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 */
%o (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
%o (Magma) [Round(3^n*Gamma(n+2/3)/Gamma(2/3)): n in [0..20]]; // _G. C. Greubel_, Mar 31 2019
%Y a(n) = A004747(n+1, 1) (first column of triangle). Cf. A051141.
%Y Cf. A000165, A001813, A047055, A047657, A084947, A084948, A084949.
%Y Cf. A029768, A034724, A049308.
%Y Cf. A024396, A193534.
%Y Cf. A225470, A290596 (first columns).
%Y Subsequence of A007661.
%K nonn,easy
%O 0,2
%A Joe Keane (jgk(AT)jgk.org)
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