|
%I #45 Mar 02 2018 02:32:36
%S 1,30,2310,204204,19122246,1848483780,182327718300,18236779032600,
%T 1842826521244230,187679234340049620,19232182592635611060,
%U 1980665038436368775400,204826599735691440534300,21255328931341321610645544,2212241139727064219063537016
%N a(n) = (6n)!n!/((3n)!(2n)!^2).
%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 = 3, 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 A211419 (a = 3, b = 2), A211420 (a = 4, b = 1) and A211421 (a = 4, b = 3) and A061163 (a = 5, b = 1). 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
%D M. Kontsevich and D. Zagier, Periods, in Mathematics Unlimited - 2001 and Beyond, Springer, Berlin, 2001, pp. 771-808.
%D R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.
%H Harry J. Smith, <a href="/A061162/b061162.txt">Table of n, a(n) for n = 0..100</a>
%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:0709.1977v1 [math.NT], J. London Math. Soc., Vol. 79, Issue 2, (2009) 422-444.
%H W. Mlotkowski and K. A. Penson, <a href="http://arxiv.org/abs/1309.0595">Probability distributions with binomial moments</a>, arXiv preprint arXiv:1309.0595 [math.PR], 2013.
%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 a(n) ~ 1/2*Pi^(-1/2)*n^(-1/2)*2^(2*n)*3^(3*n)*{1 - 1/72*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Jun 11 2002
%F n*(2*n-1)*a(n) -6*(6*n-1)*(6*n-5)*a(n-1)=0. - _R. J. Mathar_, Oct 26 2014
%F From _Peter Bala_, Aug 21 2016: (Start)
%F a(n) = Sum_{k = 0..2*n} binomial(6*n, k)*binomial(4*n - k - 1, 2*n - k).
%F a(n) = Sum_{k = 0..n} binomial(8*n, 2*n - 2*k)*binomial(2*n + k - 1, k).
%F O.g.f. A(x) = Hypergeom([5/6, 1/6], [1/2], 108*x).
%F a(n) = [x^(2*n)] H(x)^n, where H(x) = (1 + x)^6/(1 - x)^2. Cf. A091496 and A262732. It follows that the o.g.f. A(x) for this sequence is the diagonal of the bivariate rational generating function 1/2*( 1/(1 - t*H(sqrt(x))) + 1/(1 - t*H(-sqrt(x))) ) and hence is algebraic by Stanley 1999, Theorem 6.33, p. 197.
%F Let G(x) = 1/x * series reversion( x*(1 - x)/(1 + x)^3 ) = 1 + 4*x + 23*x^2 + 156*x^3 + 1162*x^4 + ..., essentially the o.g.f. for A007297. Then A(x^2) equals the even part of 1 + x*d/dx(Log(G(x)).
%F exp(Sum_{n >= 1} a(n)*x^n/n) = F(x), where F(x) = 1 + 30*x + 1605*x^2 + 107218*x^3 + 8043114*x^4 + 647773116*x^5 + 54730094637*x^6 + ... has integer coefficients since F(x^2) = G(x)*G(-x). Furthermore, F(x)^(1/6) = 1 + 5*x + 205*x^2 + 12328*x^3 + 874444*x^4 + 68022261*x^5 + 5613007167*x^6 + ... appears to have all integer coefficients. (End)
%F a(n) is the n-th moment of the positive weight function w(x) on x = (0,108), i.e., in Maple notation: a(n) = int(x^n*w(x), x=0..108), n=0,1,..., where w(x) = sqrt(3)*(1 + sqrt(1 - x/108))^(2/3)/(12*2^(1/3)*Pi*x^(5/6)*sqrt(1 - x/108)) + 2^(4/3) *sqrt(3)/(864*Pi*x^(1/6)*(1 + sqrt(1 - x/108))^(2/3)*sqrt(1 - x/108)). The weight function w(x) is singular at x=0 and at x=108 and is the solution of the Hausdorff moment problem. This solution is unique. - _Karol A. Penson_, Mar 01 2018
%p A061162 := n->(6*n)!*n!/((3*n)!*(2*n)!^2);
%t a[n_] := 16^n Gamma[3 n + 1/2]/(Gamma[n + 1/2] Gamma[2 n + 1]);
%t Table[a[n], {n, 0, 14}] (* _Peter Luschny_, Mar 01 2018 *)
%o (PARI) { for (n=0, 100, write("b061162.txt", n, " ", (6*n)!*n!/((3*n)!*(2*n)!^2)) ) } \\ _Harry J. Smith_, Jul 18 2009
%Y Cf. A061163, A061164, A007297, A091527, A262732.
%Y See also A091527, A211419, A211420, A211421.
%K easy,nonn
%O 0,2
%A _Richard Stanley_, Apr 17 2001
|
