%I #42 Nov 18 2022 11:40:52
%S 1,12,308,8976,276276,8767512,283728368,9307523160,308397041460,
%T 10297160887440,345907798472808,11677249143964768,395812039652176368,
%U 13462467839121604380,459228293024549285160,15704603616059963888976,538244114669755948787508,18482918763808824442733616
%N a(n) = (6*n)!*(4*n)!/((2*n)!*(3*n)!*(5*n)!).
%C From Bober's reference, s. Theorem 1.2: Eq.(9) for a=3, b=2 gives a(n).
%H J. W. Bober, <a href="https://arxiv.org/abs/0709.1977">Factorial ratios, hypergeometric series, and a family of step functions</a>, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc. (2) 79 (2009), 422-444.
%F O.g.f.: hypergeom([1/6,1/4,1/2,3/4,5/6],[1/5,2/5,3/5,4/5],110592*z/3125).
%F a(n) ~ sqrt(10)*sqrt(1/n)*(32248627200000*n^4 - 1657221120000*n^3 + 42581376000*n^2 + 12688032960*n - 680136331)*3125^(-n)*110592^n/(161243136000000*n^4*sqrt(Pi)).
%F Integral representation as the n-th moment of the positive function V(x) on x = (0, 110592/3125), i.e. in Maple notation: a(n) = int(x^n*V(x),x=0..110592/3125), where V(x) = 2^(1/3)*hypergeom([1/6, 11/30, 17/30, 23/30, 29/30], [1/3, 5/12, 2/3, 11/12], 3125*x*(1/110592))/(12*Pi*x^(5/6))+sqrt(2)*hypergeom([1/4, 9/20, 13/20, 17/20, 21/20], [5/12, 1/2, 3/4, 13/12], 3125*x*(1/110592))/(16*Pi*x^(3/4)) +cos((1/5)*Pi)*cos(2*Pi*(1/5))*hypergeom([1/2, 7/10, 9/10, 11/10, 13/10], [2/3, 3/4, 5/4, 4/3], 3125*x*(1/110592))/(16*Pi*sqrt(x))+11*sqrt(2)*hypergeom([3/4, 19/20, 23/20, 27/20, 31/20], [11/12, 5/4, 3/2, 19/12], 3125*x*(1/110592))/ (2048*Pi*x^(1/4))+247*2^(2/3)*hypergeom([5/6, 31/30, 37/30, 43/30, 49/30], [13/12, 4/3, 19/12, 5/3], 3125*x*(1/110592))/(110592*Pi*x^(1/6)). The function V(x) is singular at both edges of its support and is U-shaped. The function V(x) is unique as it is the solution of the Hausdorff moment problem.
%F a(n) = (2^(10*n)*Gamma(1/2 + 2*n)*Gamma(1/2 + 3*n))/(Pi*Gamma(1 + 5*n)). - _Peter Luschny_, May 07 2018
%F D-finite with recurrence 5*n*(5*n-4)*(5*n-3)*(5*n-2)*(5*n-1)*a(n) -96*(6*n-5)*(6*n-1)*(4*n-3)*(2*n-1)*(4*n-1)*a(n-1)=0. - _R. J. Mathar_, Jul 27 2022
%F From _Karol A. Penson_, Nov 18 2022: (Start)
%F O.g.f. = h(z) satisfies the following algebraic equation of order 16:
%F 268435456*z - 759375 + (-8657043456*z + 32400000)*h(z) + (113508352000*z - 590625000)*h(z)^2 + (-737869824000*z + 5850000000)*h(z)^3 + (1099511627776*z^2 + 2023598080000*z - 32039062500)*h(z)^4 + (-22265110462464*z^2 + 2487091200000*z + 72500000000)*h(z)^5 + (160030481448960*z^2 - 30842368000000*z + 172265625000)*h(z)^6 + (-234827336908800*z^2 + 57507840000000*z - 1437500000000)*h(z)^7 + (-2334117593088000*z^2 + 82630400000000*z + 1970214843750)*h(z)^8 + (8806025134080000*z^2 - 497664000000000*z + 7031250000000)*h(z)^9 + (25649407252758528*z^3 + 2525390438400000*z^2 + 659840000000000*z - 21240234375000)*h(z)^10 + (519400496868360192*z^3 - 44030125670400000*z^2 + 1244160000000000*z - 11718750000000)*h(z)^11 + (405781638178406400*z^3 + 80263249920000000*z^2 - 5508000000000000*z + 82397460937500)*h(z)^12 + (6763027302973440000*z^3 - 573308928000000000*z^2 + 16200000000000000*z - 152587890625000)*h(z)^14 + (149587343098087735296*z^4 - 16907568257433600000*z^3 + 716636160000000000*z^2 - 13500000000000000*z + 95367431640625)*h(z)^16=0. (End)
%p seq((6*n)!*(4*n)!/((2*n)!*(3*n)!*(5*n)!),n=0..17);
%p # Alternative (to illustrate the constant C):
%p k := [10,15,30,45,50]/60: j := [12,24,36,48,60]/60: C := 110592/3125:
%p a := n -> C^n*mul(pochhammer(k[i],n)/pochhammer(j[i],n), i=1..5): # _Peter Luschny_, May 07 2018
%t a[n_] := (2^(10 n) Gamma[1/2 + 2 n] Gamma[1/2 + 3 n])/(Pi Gamma[1 + 5 n]);
%t Table[a[n], {n, 0, 17}] (* _Peter Luschny_, May 07 2018 *)
%t Table[((6n)!(4n)!)/((2n)!(3n)!(5n)!),{n,0,20}] (* _Harvey P. Dale_, Jul 26 2019 *)
%Y Cf. A295431.
%K nonn
%O 0,2
%A _Karol A. Penson_, May 07 2018