A275607 a(n) = 2*12^n*Gamma(n+1/2)*(n+1)/(sqrt(Pi)*Gamma(n+3)).

1, 4, 27, 216, 1890, 17496, 168399, 1667952, 16888014, 173997720, 1818276174, 19225409616, 205299909828, 2210922105840, 23984556773175, 261854925711840, 2874948871877910, 31722346066169880, 351589335566716170, 3912422681494285200, 43694647856506630620, 489597172255515289680

Offset: 0

Author

Karol A. Penson, Nov 14 2016

Keywords

• nonn

Comments

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.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..925
• Simeon T. Stefanov, Counting fixed points free vector fields on B^2, arXiv:1807.03714 [math.GT], 2018.
• K. Szymanski, B. Collins, T. Szarek and K. Zyczkowski, Convex set of quantum states with positive partial transpose analysed by hit and run algorithm, arXiv:1611.01194 [quant-ph], 2016.

Crossrefs

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

Programs

Maple

a := n -> (2^(2*n+1)*3^n*(n+1)*GAMMA(n+1/2))/(sqrt(Pi)*GAMMA(n+3)):
seq(a(n), n=0..21); # Peter Luschny, Nov 14 2016

Mathematica

g[z_] :=  E^z (BesselI[0,z] - (1-1/z) BesselI[1,z])
Table[CoefficientList[2/3 Series[g[6z], {z,0,21}],z]] Range[0, 21]! //Flatten (* Peter Luschny, Nov 14 2016 *)
Table[ 2*12^n*(n + 1)*Gamma[n + 1/2]/(Sqrt[Pi]*Gamma[n + 3]), {n,0,100}] (* G. C. Greubel, Jan 13 2017 *)

PARI

a(n)=2*12^n*gamma(n+1/2)*(n+1)\/(sqrt(Pi)*(n+2)!) \\ Charles R Greathouse IV, Nov 14 2016

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

Formula

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

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;

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.

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

a(n) ~ 2^(2*n+1)*3^n/(sqrt(Pi)*n^(3/2)). - Ilya Gutkovskiy, Nov 14 2016

a(n) = 2*3^n*binomial(2n+1, n-1)*(n+1)/(2n^2+n). - Charles R Greathouse IV, Nov 14 2016