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A211420
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a(n) = (8*n)!*n!/((4*n)!*(3*n)!*(2*n)!).
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6
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1, 140, 60060, 29745716, 15628090140, 8480843582640, 4697400936504900, 2638798257262351800, 1497753729733989900060, 856840435680656569701776, 493243073668546377605912560, 285369375758780754651194529300, 165789876049841088844342275759300
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OFFSET
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0,2
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COMMENTS
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This sequence is the particular case a = 4, b = 1 of the following result (see Bober, Theorem 1.2): let a, b be nonnegative integers with a > b and GCD(a,b) = 1. Then (2*a*n)!*(b*n)!/((a*n)!*(2*b*n)!*((a-b)*n)!) is an integer for all integer n >= 0. Other cases include A061162 (a = 3, b = 1), A211419(a = 3, b = 2), A211421(a = 4, b = 3) and A061163 (a = 5, b = 1).
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REFERENCES
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R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.
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LINKS
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FORMULA
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The o.g.f. Sum_{n >= 1} a(n)*z^n is algebraic over the field of rational functions Q(z) (see Rodriguez-Villegas).
D-finite with recurrence: 3*(3*n-1)*(2*n-1)*(3*n-2)*n*a(n) - 8*(8*n-3)*(8*n-1)*(8*n-7)*(8*n-5)*a(n-1) = 0. - Georg Fischer, Nov 30 2022
a(n) = Sum_{k = 0..3*n} binomial(8*n, k)*binomial(5*n-k-1, 3*n-k).
a(n) = [x^(3*n)] F(x)^n, where F(x) = (1 + x)^8/(1 - x)^2.
It follows that the o.g.f. A(x) for this sequence is the diagonal of the bivariate rational generating function 1/3*( 1/(1 - t*F(x^(1/3))) + 1/(1 - t*F(w*x^(1/3))) + 1/(1 - t*F(w^2*x^(1/3))) ), where w = exp(2*Pi*i/3), and hence A(x), as stated above, is algebraic over Q(x) by Stanley 1999, Theorem 6.33, p. 197. (End)
O.g.f.: hypergeometric4F3([1/8, 3/8, 5/8, 7/8], [1/3, 1/2, 2/3], (2^14*z)/27). (O.g.f.(z))^2 satisfies the algebraic equation of order 16, in which the powers of (o.g.f.(z))^2 are multiplied by polynomials p(n, z) with integer coefficients, in the form: Sum_{n = 0..16} p(n, z) * (o.g.f.(z))^(2*n) = 0.
Here is the list of orders, in the variable z, of all polynomials p(n, z) for n = 0..16: 8,8,8,8,9,9,9,9,10,10,10,11,11,11,11,11,12. For example p(14, z) = 6*(2^13*z + 31*27)*(2^14*z - 27)^10. (End)
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MAPLE
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a := n -> (2^(6*n)*GAMMA(4*n + 1/2))/(GAMMA(n + 1/2)*GAMMA(3*n + 1)):
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MATHEMATICA
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Table[ 2^(6*n) * Gamma[4*n + 1/2] / (Gamma[n + 1/2] * Gamma[3*n + 1]), {n, 0, 12}] (* James C. McMahon, Feb 24 2024 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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