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%I #29 Nov 15 2020 03:11:34
%S 1,6,360,60480,19958400,10897286400,8892185702400,10137091700736000,
%T 15388105201717248000,30006805143348633600000,
%U 73096577329197271449600000,217535414131691079834009600000
%N a(n) = (3*n)!/n!.
%C Also a(n) = (((n)!)^2)*A006480(n). [corrected by _Johannes W. Meijer_, Mar 02 2009]
%C a(n) is the number of ways to partition the set {1,2,...,3n} into n blocks of size 3 and then linearly order the elements within each block. - _Geoffrey Critzer_, Dec 30 2012
%H Harry J. Smith, <a href="/A064350/b064350.txt">Table of n, a(n) for n = 0..70</a>
%H Karol A. Penson and Allan I. Solomon, <a href="https://doi.org/10.1142/9789812777850_0066">Coherent states from combinatorial sequences</a>, in: E. Kapuscik and A. Horzela (eds.), Quantum theory and symmetries, World Scientific, 2002, pp. 527-530; <a href="https://arxiv.org/abs/quant-ph/0111151">arXiv preprint</a>, arXiv:quant-ph/0111151, 2001.
%F Integral representation as n-th moment of a positive function on a positive half-axis, in Maple notation: a(n)=int(x^n*BesselK(1/3, 2*sqrt(x/27))/(3*Pi*sqrt(x)), x=0..infinity), n=0, 1, ...
%F A recursive formula: a(n) = (27 * (n - 1)^2 + 27 * (n - 1) + 6) * a(n - 1) with a(0) = 1. An explicit formula following from the recursion equation: a(n) = (3/2)*27^n*GAMMA(n+2/3)*GAMMA(n+1/3)/(Pi*3^(1/2)). - _Thomas Wieder_, Nov 15 2004
%F E.g.f.: (of aerated sequence) 2*cos(arcsin((3*sqrt(3)*x/2)/3))/sqrt(4-27*x^2). - _Paul Barry_, Jul 27 2010
%F E.g.f.: (with interpolated zeros) exp(x^3). - _Geoffrey Critzer_, Dec 30 2012
%F Sum_{n>=1} 1/a(n) = A248759. - _Amiram Eldar_, Nov 15 2020
%t Table[(3n)!/n!,{n,0,20}] (* _Geoffrey Critzer_, Dec 30 2012 *)
%o (PARI) { t=f=1; for (n=0, 70, if (n, t*=3*n*(3*n - 1)*(3*n - 2); f*=n); write("b064350.txt", n, " ", t/f) ) } \\ _Harry J. Smith_, Sep 12 2009
%Y Cf. A006480, A248759.
%Y From _Johannes W. Meijer_, Mar 07 2009: (Start)
%Y Equals A001525*3!
%Y Equals row sums of A157704 and A157705. (End)
%K nonn
%O 0,2
%A _Karol A. Penson_, Sep 18 2001
%E a(11) from _Harry J. Smith_, Sep 12 2009
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