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A304126
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a(n) = (6*n)!*(4*n)!/((2*n)!*(3*n)!*(5*n)!).
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5
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1, 12, 308, 8976, 276276, 8767512, 283728368, 9307523160, 308397041460, 10297160887440, 345907798472808, 11677249143964768, 395812039652176368, 13462467839121604380, 459228293024549285160, 15704603616059963888976, 538244114669755948787508, 18482918763808824442733616
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OFFSET
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0,2
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COMMENTS
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From Bober's reference, s. Theorem 1.2: Eq.(9) for a=3, b=2 gives a(n).
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LINKS
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FORMULA
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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).
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)).
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.
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
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
O.g.f. = h(z) satisfies the following algebraic equation of order 16:
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)
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MAPLE
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seq((6*n)!*(4*n)!/((2*n)!*(3*n)!*(5*n)!), n=0..17);
# Alternative (to illustrate the constant C):
k := [10, 15, 30, 45, 50]/60: j := [12, 24, 36, 48, 60]/60: C := 110592/3125:
a := n -> C^n*mul(pochhammer(k[i], n)/pochhammer(j[i], n), i=1..5): # Peter Luschny, May 07 2018
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MATHEMATICA
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a[n_] := (2^(10 n) Gamma[1/2 + 2 n] Gamma[1/2 + 3 n])/(Pi Gamma[1 + 5 n]);
Table[((6n)!(4n)!)/((2n)!(3n)!(5n)!), {n, 0, 20}] (* Harvey P. Dale, Jul 26 2019 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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