A051578 a(n) = (2*n+4)!!/4!!, related to A000165 (even double factorials).

1, 6, 48, 480, 5760, 80640, 1290240, 23224320, 464486400, 10218700800, 245248819200, 6376469299200, 178541140377600, 5356234211328000, 171399494762496000, 5827582821924864000, 209792981589295104000, 7972133300393213952000, 318885332015728558080000

Offset: 0

Author

Wolfdieter Lang

Keywords

• easy
• nonn

Comments

Row m=4 of the array A(3; m,n) := (2*n+m)!!/m!!, m >= 0, n >= 0.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..400
• INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 521.
• A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.

Crossrefs

Cf. A000165, A001147(n+1), A002866(n+1), A051577 (rows m=0..3), A051579, A051580, A051581, A051582, A051583.

Cf. A052587 (essentially the same).

Programs

GAP

List([0..20], n-> 2^(n-1)*Factorial(n+2) ); # G. C. Greubel, Nov 11 2019

Magma

[2^(n-1)*Factorial(n+2): n in [0..20]]; // G. C. Greubel, Nov 11 2019

Maple

a:= proc(n) option remember; `if`(n=0, 1, 2*(n+2)*a(n-1)) end:
seq(a(n), n=0..20);  # Alois P. Heinz, Apr 29 2019
seq(2^(n-1)*(n+2)!, n=0..20); # G. C. Greubel, Nov 11 2019

Mathematica

Table[2^(n-1)(n+2)!, {n,0,20}] (* Jean-François Alcover, Oct 05 2019 *)
Table[(2n+4)!!/8,{n,0,20}] (* Harvey P. Dale, Apr 06 2023 *)

PARI

vector(21, n, 2^(n-2)*(n+1)! ) \\ G. C. Greubel, Nov 11 2019

PARI

apply( {A051578(n)=(n+2)!<<(n-1)}, [0..18]) \\ M. F. Hasler, Nov 10 2024

Sage

[2^(n-1)*factorial(n+2) for n in (0..20)] # G. C. Greubel, Nov 11 2019

Formula

a(n) = (2*n+4)!!/4!!.

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

a(n) ~ 2^(-1/2)*Pi^(1/2)*n^(5/2)*2^n*e^-n*n^n*{1 + 37/12*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Nov 23 2001

a(n) = (n+2)!*2^(n-1). - Zerinvary Lajos, Sep 23 2006. [corrected by Gary Detlefs, Apr 29 2019]

a(n) = 2^n*A001710(n+2). - R. J. Mathar, Feb 22 2008

From Peter Bala, May 26 2017: (Start)

a(n+1) = (2*n + 6)*a(n) with a(0) = 1.

O.g.f. satisfies the Riccati differential equation 2*x^2*A(x)' = (1 - 6*x)*A(x) - 1 with A(0) = 1.

G.f. as an S-fraction: A(x) = 1/(1 - 6*x/(1 - 2*x/(1 - 8*x/(1 - 4*x/(1 - 10*x/(1 - 6*x/(1 - ... - (2*n + 4)*x/(1 - 2*n*x/(1 - ...))))))))) (by Stokes 1982).

Reciprocal as an S-fraction: 1/A(x) = 1/(1 + 6*x/(1 - 8*x/(1 - 2*x/(1 - 10*x/(1 - 4*x/(1 - 12*x/(1 - 6*x/(1 - ... - (2*n + 6)*x/(1 - 2*n*x/(1 - ...)))))))))). (End)

From Amiram Eldar, Dec 11 2022: (Start)

Sum_{n>=0} 1/a(n) = 8*sqrt(e) - 12.

Sum_{n>=0} (-1)^n/a(n) = 8/sqrt(e) - 4. (End)

a(n) = A052587(n+2) for n > 0. - M. F. Hasler, Nov 10 2024