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%I #46 Feb 23 2024 06:36:41
%S 1,630,1385670,3528923580,9540949030470,26651569523959380,
%T 75998432812419471900,219813190240007470094520,
%U 642409325786050322446410310,1892390644737640220059489996260
%N a(n) = (10n)!*n!/((5n)!*(4n)!*(2n)!).
%C According to page 781 of the cited reference the generating function F(x) for a(n) is algebraic but not obviously so and the minimal polynomial satisfied by F(x) is quite large.
%C This sequence is the particular case a = 5, 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) and A211420(a = 4, b = 1) and A211421 (a = 4, b = 3). The o.g.f. Sum_{n >= 1} a(n)*z^n is algebraic over the field of rational functions Q(z) (see Rodriguez-Villegas). - _Peter Bala_, Apr 10 2012
%C Continuing the comment above: This is case n = 4 of the array of sequences
%C A(n, k) = 4^(n*k)*(Gamma((n + 1)*k + 1/2)/Gamma(k + 1/2)) / Gamma(n * k + 1). See the cross-references for other cases. - _Peter Luschny_, Feb 21 2024
%D M. Kontsevich and D. Zagier, Periods, in Mathematics Unlimited - 2001 and Beyond, Springer, Berlin, 2001, pp. 771-808.
%H J. W. Bober, <a href="http://arxiv.org/abs/0709.1977">Factorial ratios, hypergeometric series, and a family of step functions</a>, 2007, arXiv:0709.1977 [math.NT], 2007; Jour. of the London Math. Soc., Vol. 79, Issue 2, 422-444.
%H F. Rodriguez-Villegas, <a href="http://arxiv.org/abs/math/0701362">Integral ratios of factorials and algebraic hypergeometric functions</a>, arXiv:math/0701362 [math.NT], 2007.
%F n*(4*n-3)*(2*n-1)*(4*n-1)*a(n) -10*(10*n-9)*(10*n-7)*(10*n-3)*(10*n-1)*a(n-1)=0. - _R. J. Mathar_, Oct 26 2014
%F O.g.f. is a generalized hypergeometric function 4F3([1/10, 3/10, 7/10, 9/10], [1/4, 1/2, 3/4], 5^5*z). - _Karol A. Penson_, Apr 13 2022
%F From _Karol A. Penson_, Feb 21 2024: (Start)
%F (O.g.f.(z))^2 satisfies the algebraic equation of order 15, 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..15} p(n, z)*(O.g.f.(z))^(2*n) = 0.
%F Here is the list of orders, in the variable z, of all polynomials p(n, z) for n=0..15: 9,9,9,9,9,10,10,10,10,10,10,11,11,11,11,11,12. For example p(15, z) = 2^50*(5^5*z-1)^12. (End)
%p A061163 := n->(10*n)!*n!/((5*n)!*(4*n)!*(2*n)!);
%p # Alternative:
%p A := (n, k) -> 4^(n*k)*(GAMMA((n + 1)*k + 1/2)/GAMMA(k + 1/2))/GAMMA(n*k + 1):
%p seq(A(4, k), k = 0..9); # _Peter Luschny_, Feb 21 2024
%t Table[(10n)! n!/((5n)!(4n)!(2n)!),{n,0,10}] (* _Harvey P. Dale_, Oct 24 2022 *)
%Y Cf. A061164, A211419, A211421.
%Y Cf. A000012 (n=0), A001448 (n=1), A061162 (n=2), A211420 (n=3), this sequence (n=4).
%K easy,nonn
%O 0,2
%A _Richard Stanley_, Apr 17 2001
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