A011781 Sextuple factorial numbers: a(n) = Product_{k=0..n-1} (6*k+3).

1, 3, 27, 405, 8505, 229635, 7577955, 295540245, 13299311025, 678264862275, 38661097149675, 2435649120429525, 168059789309637225, 12604484198222791875, 1020963220056046141875, 88823800144876014343125, 8260613413473469333910625, 817800727933873464057151875

Offset: 0

Author

Lee D. Killough (killough(AT)wagner.convex.com)

Keywords

• nonn
• easy

Comments

Total number of Eulerian circuits in rooted labeled multigraphs with n edges. - Valery A. Liskovets, Apr 07 2002

Number of self-avoiding planar walks starting at (0,0), ending at (n,0), remaining in the east quadrant {(x,y): x >= |y|} and using steps (0,1), (0,-1), (1,1), (-1,-1), and (1,0). - Alois P. Heinz, Oct 13 2016

Examples

			G.f. = 1 + 3*x + 27*x^2 + 405*x^3 + 8505*x^4 + 229635*x^5 + 7577955*x^6 + ...
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• Fatemeh Bagherzadeh, M. Bremner, and S. Madariaga, Jordan Trialgebras and Post-Jordan Algebras, arXiv:1611.01214 [math.RA], 2016.
• Murray Bremner and Martin Markl, Distributive laws between the Three Graces, arXiv:1809.08191 [math.AT], 2018.
• Bodo Lass, Démonstration combinatoire de la formule de Harer-Zagier, (A combinatorial proof of the Harer-Zagier formula) C. R. Acad. Sci. Paris, Serie I, 333(3) (2001), 155-160.
• Bodo Lass, Démonstration combinatoire de la formule de Harer-Zagier, (A combinatorial proof of the Harer-Zagier formula) C. R. Acad. Sci. Paris, Serie I, Vol. 333, No. 3 (2001), pp. 155-160; alternative link.
• Valery Liskovets, A Note on the Total Number of Double Eulerian Circuits in Multigraphs , Journal of Integer Sequences, Vol. 5 (2002), Article 02.2.5.

Crossrefs

Cf. A001147, A035309, A047657, A048994, A049308, A069736, A273464, A277358.

Programs

GAP

F:=Factorial;; List([0..20], n-> (3/2)^n*(F(2*n)/F(n)) ); # G. C. Greubel, Aug 20 2019

Magma

[(3/2)^n*Factorial(2*n)/Factorial(n):n in [0..20]]; // Vincenzo Librandi, May 09 2012

Mathematica

Table[Product[6k+3,{k,0,n-1}],{n,0,20}] (* or *) Table[6^(n-1) Pochhammer[ 1/2,n-1],{n,21}] (* Harvey P. Dale, May 09 2012 *)
Table[6^n*Pochhammer[1/2, n], {n,0,20}] (* G. C. Greubel, Aug 20 2019 *)

PARI

{a(n) = if( n<0, (-1)^n / a(-n), (3/2)^n * (2*n)! / n!)}; /* Michael Somos, Feb 10 2002, revised and extended Michael Somos, Jan 06 2017 */

Sage

[6^n*rising_factorial(1/2, n) for n in (0..20)] # G. C. Greubel, Aug 20 2019

Formula

E.g.f.: (1-6*x)^(-1/2).

a(n) = 3^n*(2*n-1)!!.

G.f.: 1/(1-3*x/(1-6*x/(1-9*x/(1-12*x/(1-15*x/(1-18*x/(1-21*x/(1-24*x/(1-... (continued fraction). - Philippe Deléham, Jan 08 2012

a(n) = (-3)^n*Sum_{k=0..n} 2^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.: T(0), where T(k) = 1 - 3*x*(k+1)/( 3*x*(k+1) - 1/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 24 2013

a(n) = 6^n * gamma(n + 1/2) / sqrt(Pi). - Daniel Suteu, Jan 06 2017

0 = a(n)*(+6*a(n+1) - a(n+2)) + a(n+1)*(+a(n+1)) and a(n) = (-1)^n / a(-n) for all n in Z. - Michael Somos, Jan 06 2017

D-finite with recurrence: a(n) +3*(-2*n+1)*a(n-1)=0. - R. J. Mathar, Jan 20 2018

From Amiram Eldar, Feb 11 2022: (Start)

Sum_{n>=0} 1/a(n) = 1 + exp(1/6)*sqrt(Pi/6)*erf(1/sqrt(6)), where erf is the error function.

Sum_{n>=0} (-1)^n/a(n) = 1 - exp(-1/6)*sqrt(Pi/6)*erfi(1/sqrt(6)), where erfi is the imaginary error function. (End)