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A066318 Number of necklaces with n labeled beads of 2 colors. 5

%I

%S 2,4,16,96,768,7680,92160,1290240,20643840,371589120,7431782400,

%T 163499212800,3923981107200,102023508787200,2856658246041600,

%U 85699747381248000,2742391916199936000,93241325150797824000,3356687705428721664000,127554132806291423232000

%N Number of necklaces with n labeled beads of 2 colors.

%C 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

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

%H Vincenzo Librandi, <a href="/A066318/b066318.txt">Table of n, a(n) for n = 1..400</a>

%H Alexsandar Petojevic, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL5/Petojevic/petojevic5.html">The Function vM_m(s; a; z) and Some Well-Known Sequences</a>, Journal of Integer Sequences, Vol. 5 (2002), Article 02.1.7

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Normal_distribution">Normal distribution</a>, formula for E(|x|^p).

%H <a href="/index/Ne#necklaces">Index entries for sequences related to necklaces</a>

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

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

%F 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

%F a(n) = 2*(n-1)*a(n-1). - _R. J. Mathar_, Sep 10 2012

%F 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

%F a(n) = 2 * (2*n-2)!! = 2 * A000165(n-1). - _Alois P. Heinz_, Jun 22 2017

%F 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

%p seq(count(Permutation(n))*count(Subset(n+1)),n=0..17); # _Zerinvary Lajos_, Oct 16 2006

%p with(combstruct):A:=[N,{N=Cycle(Union(Z$2))},labeled]: seq(count(A,size=n),n=1..18); # _Zerinvary Lajos_, Oct 07 2007

%p # alternative Maple program:

%p a:= n-> 2*doublefactorial(2*n-2):

%p seq(a(n), n=1..20); # _Alois P. Heinz_, Jun 22 2017

%t mx = 18; Rest[ Range[0, mx]! CoefficientList[ Series[ Log[1/(1 - 2 x)], {x, 0, mx}], x]] (* _Robert G. Wilson v_, Sep 22 2011 *)

%t Table[(n-1)!*2^n,{n,20}] (* _Harvey P. Dale_, Dec 15 2011 *)

%o (MAGMA) [Factorial(n-1)*2^n: n in [1..20]]; // _Vincenzo Librandi_, Sep 23 2011

%o (PARI) apply( A066318=n->(n-1)!<<n ,[1..18]) \\ _M. F. Hasler_, Jan 15 2017

%o (GAP) a_n:=List([1..10], n->Factorial(n-1)*2^n); # _Stefano Spezia_, Nov 17 2018

%o (Python) import math

%o for n in range(1,10): print(math.factorial(n-1)*2**n, end=', ') # _Stefano Spezia_, Nov 17 2018

%o (Maxima) a(n):=(n-1)!*2^n$ makelist(a(n), n, 1, 10); /* _Stefano Spezia_, Nov 21 2018 */

%o (Sage) [2^n*factorial(n-1) for n in (1..20)] # _G. C. Greubel_, Nov 21 2018

%Y Apart from initial term, same as A032184.

%Y Cf. A000165, A000796, A001147, A019727.

%K nonn

%O 1,1

%A _Christian G. Bower_, Dec 13 2001

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Last modified January 19 18:27 EST 2019. Contains 319309 sequences. (Running on oeis4.)