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%I #33 Sep 08 2022 08:44:52
%S 1,9,135,2835,76545,2525985,98513415,4433103675,226088287425,
%T 12887032383225,811883040143175,56019929769879075,4201494732740930625,
%U 340321073352015380625,29607933381625338114375,2753537804491156444636875,272600242644624488019050625
%N a(n) is the n-th sextic factorial number divided by 3.
%H G. C. Greubel, <a href="/A034723/b034723.txt">Table of n, a(n) for n = 1..345</a>
%F 3*a(n) = (6*n-3)(!^6) = Product_{j=1..n} (6*j-3) = 3^n*A001147(n) = 3^n*(2*n)!/(2^n*n!).
%F E.g.f.: (-1 + (1-6*x)^(-1/2))/3.
%F a(n) = 2*(3/2)^(n-1)*(n+1)!*C(n), where C(n) = A000108(n). - _G. C. Greubel_, Nov 11 2019
%F D-finite with recurrence: a(n) + 3*(-2*n+1)*a(n-1) = 0. - _R. J. Mathar_, Feb 24 2020
%F From _Amiram Eldar_, Jan 08 2022: (Start)
%F Sum_{n>=1} 1/a(n) = e^(1/6)*sqrt(3*Pi/2)*erf(1/sqrt(6)), where erf is the error function.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = e^(-1/6)*sqrt(3*Pi/2)*erfi(1/sqrt(6)), where erfi is the imaginary error function. (End)
%p seq(3^(n-1)*(2*n)!/(2^n*n!), n=1..20); # _G. C. Greubel_, Nov 11 2019
%t Table[3^(n-1)*(2*n)!/(2^n*n!), {n,20}] (* _G. C. Greubel_, Nov 11 2019 *)
%o (PARI) a(n) = prod(j=1, n, 6*j-3)/3; \\ _Michel Marcus_, Mar 13 2019
%o (Magma) F:=Factorial; [3^(n-1)*F(2*n)/(2^n*F(n)): n in [1..20]]; // _G. C. Greubel_, Nov 11 2019
%o (Sage) f=factorial; [3^(n-1)*f(2*n)/(2^n*f(n)) for n in (1..20)] # _G. C. Greubel_, Nov 11 2019
%o (GAP) F:=Factorial;; List([1..20], n-> 3^(n-1)*F(2*n)/(2^n*F(n))); # _G. C. Greubel_, Nov 11 2019
%Y Cf. A000108, A001147, A008542, A034689.
%K easy,nonn
%O 1,2
%A _Wolfdieter Lang_
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