A275607 - OEIS

%I #44 Oct 15 2018 22:16:13

%S 1,4,27,216,1890,17496,168399,1667952,16888014,173997720,1818276174,

%T 19225409616,205299909828,2210922105840,23984556773175,

%U 261854925711840,2874948871877910,31722346066169880,351589335566716170,3912422681494285200,43694647856506630620,489597172255515289680

%N a(n) = 2*12^n*Gamma(n+1/2)*(n+1)/(sqrt(Pi)*Gamma(n+3)).

%C In reference of K. Szymanski et al. the function g(x) from the Eq.(4.6) satisfies the equality g(x/4)/4 = W(x) where W(x) is the weight function of the integral representation, see below.

%H G. C. Greubel, <a href="/A275607/b275607.txt">Table of n, a(n) for n = 0..925</a>

%H Simeon T. Stefanov, <a href="https://arxiv.org/abs/1807.03714">Counting fixed points free vector fields on B^2</a>, arXiv:1807.03714 [math.GT], 2018.

%H K. Szymanski, B. Collins, T. Szarek and K. Zyczkowski, <a href="https://arxiv.org/abs/1611.01194">Convex set of quantum states with positive partial transpose analysed by hit and run algorithm</a>, arXiv:1611.01194 [quant-ph], 2016.

%F O.g.f: (1/54)*(1-(6*z+1)*sqrt(1-12*z))/z^2;

%F E.g.f.(in Maple notation): (1/9)*exp(6*z)*(6*z*(BesselI(0,6*z)-BesselI(1,6*z))+ BesselI(1,6*z))/z;

%F Recurrence: (-12*n^2-54*n-54)*a(n+1)+(n^2+6*n+8)*a(n+2)=0, n=0,1..., for the initial values a(0)=1, a(1)=4.

%F Integral representation as the n-th Hausdorff moment of the positive function W(x) on the segment x=(0,12), i.e., in Maple notation, a(n)= int(x^n*W(x),x=0..12), where W(x)=(1/27)*sqrt(12-x)*(3+(1/2)*x)/(Pi*sqrt(x)). This representation is unique.

%F a(n) ~ 2^(2*n+1)*3^n/(sqrt(Pi)*n^(3/2)). - _Ilya Gutkovskiy_, Nov 14 2016

%F a(n) = 2*3^n*binomial(2n+1, n-1)*(n+1)/(2n^2+n). - _Charles R Greathouse IV_, Nov 14 2016

%p a := n -> (2^(2*n+1)*3^n*(n+1)*GAMMA(n+1/2))/(sqrt(Pi)*GAMMA(n+3)):

%p seq(a(n), n=0..21); # _Peter Luschny_, Nov 14 2016

%t g[z_] :=  E^z (BesselI[0,z] - (1-1/z) BesselI[1,z])

%t Table[CoefficientList[2/3 Series[g[6z], {z,0,21}],z]] Range[0, 21]! //Flatten (* _Peter Luschny_, Nov 14 2016 *)

%t Table[ 2*12^n*(n + 1)*Gamma[n + 1/2]/(Sqrt[Pi]*Gamma[n + 3]), {n,0,100}] (* _G. C. Greubel_, Jan 13 2017 *)

%o (PARI) a(n)=2*12^n*gamma(n+1/2)*(n+1)\/(sqrt(Pi)*(n+2)!) \\ _Charles R Greathouse IV_, Nov 14 2016

%o (PARI) a(n)=2*3^n*binomial(2*n+1,n-1)*(n+1)/(2*n+1)/n \\ _Charles R Greathouse IV_, Nov 14 2016

%Y Cf. A000108, A002293, A002894, A000168, A000139, A000257, A004987, A005568, A000888, A004981.

%K nonn

%O 0,2

%A _Karol A. Penson_, Nov 14 2016