OFFSET
0,2
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.
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., 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.
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
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Karol A. Penson, Nov 14 2016
STATUS
approved