A066318 Number of necklaces with n labeled beads of 2 colors.

2, 4, 16, 96, 768, 7680, 92160, 1290240, 20643840, 371589120, 7431782400, 163499212800, 3923981107200, 102023508787200, 2856658246041600, 85699747381248000, 2742391916199936000, 93241325150797824000, 3356687705428721664000, 127554132806291423232000

Offset: 1

Author

Christian G. Bower, Dec 13 2001

Keywords

• nonn

Comments

In the normal probability distribution with mean 0 and standard deviation 1, the expected value E[|x|^(2n-1)] = a(n)/sqrt(2*Pi), while E[|x|^(2n)] = E[x^(2n)] = A001147(n). - Stanislav Sykora, Jan 15 2017

References

• F. Bergeron, G. Labelle and P. Leroux, Combinatorial Species and Tree-Like Structures, Cambridge, 1998, p. 66 (2.1.27,29).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..400
• Alexsandar Petojevic, The Function vM_m(s; a; z) and Some Well-Known Sequences, Journal of Integer Sequences, Vol. 5 (2002), Article 02.1.7.
• Wikipedia, Normal distribution, formula for E(|x|^p).
• Index entries for sequences related to necklaces.

Crossrefs

Apart from initial term, same as A032184.

Cf. A000165, A000796, A001147, A019775, A019727.

Programs

GAP

a_n:=List([1..10], n->Factorial(n-1)*2^n); # Stefano Spezia, Nov 17 2018

Magma

[Factorial(n-1)*2^n: n in [1..20]]; // Vincenzo Librandi, Sep 23 2011

Maple

with(combstruct):A:=[N,{N=Cycle(Union(Z$2))},labeled]: seq(count(A,size=n),n=1..18); # Zerinvary Lajos, Oct 07 2007
# alternative Maple program:
a:= n-> 2*doublefactorial(2*n-2):
seq(a(n), n=1..20);  # Alois P. Heinz, Jun 22 2017

Mathematica

mx = 18; Rest[ Range[0, mx]! CoefficientList[ Series[ Log[1/(1 - 2 x)], {x, 0, mx}], x]] (* Robert G. Wilson v, Sep 22 2011 *)
Table[(n-1)!*2^n,{n,20}] (* Harvey P. Dale, Dec 15 2011 *)

Maxima

a(n):=(n-1)!*2^n$ makelist(a(n), n, 1, 10);  /* Stefano Spezia, Nov 21 2018 */

PARI

apply( A066318=n->(n-1)!<M. F. Hasler, Jan 15 2017

Python

import math
for n in range(1,10): print(math.factorial(n-1)*2**n, end=', ') # Stefano Spezia, Nov 17 2018

Sage

[2^n*factorial(n-1) for n in (1..20)] # G. C. Greubel, Nov 21 2018

Formula

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

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

Let gd(x,n) = (d^n/dx^n)(exp(-(1/2)*x^2)*sqrt(2)/(2*sqrt(Pi))) = (-1)^((1/2)*n)*(x^2)^((1/2)*n)*2^(-(1/2)*n+1/2)*(exp(I*Pi*n)+1)/(4*sqrt(Pi)*GAMMA(1+(1/2)*n)) be the n-th derivative of the standard Gaussian distribution. Evaluating gd(x,n) at x=1 gives gd(1,n) = 2^(-(1/2)*n+1/2)*(exp(I*Pi*n)+1)*(-1)^((1/2)*n)/(4*sqrt(Pi)*GAMMA(1+(1/2)*n)). A066318 is the denominator of the even summands of the Taylor series expansion of the Gaussian distribution evaluated at x=1. a(n)=denom(gd(1, 2*n))/sqrt(Pi). - Stephen Crowley, May 16 2009

a(n) = 2*(n-1)*a(n-1). - R. J. Mathar, Sep 10 2012

G.f.: G(0), where G(k)= 1 + 1/(1 - 1/(1 + 1/(2*k+2)/x/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 14 2013

a(n) = 2 * (2*n-2)!! = 2 * A000165(n-1). - Alois P. Heinz, Jun 22 2017

a(n) = (sqrt(Pi)/Gamma((2*n+3)/2))*Product_{k=0..n-1} binomial(2*(n-k)+1,2). - Stefano Spezia, Nov 17 2018

From Amiram Eldar, Mar 11 2022: (Start)

Sum_{n>=1} 1/a(n) = sqrt(e)/2 (A019775).

Sum_{n>=1} (-1)^(n+1)/a(n) = 1/(2*sqrt(e)). (End)