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A034723
a(n) is the n-th sextic factorial number divided by 3.
9
1, 9, 135, 2835, 76545, 2525985, 98513415, 4433103675, 226088287425, 12887032383225, 811883040143175, 56019929769879075, 4201494732740930625, 340321073352015380625, 29607933381625338114375, 2753537804491156444636875, 272600242644624488019050625
OFFSET
1,2
LINKS
FORMULA
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!).
E.g.f.: (-1 + (1-6*x)^(-1/2))/3.
a(n) = 2*(3/2)^(n-1)*(n+1)!*C(n), where C(n) = A000108(n). - G. C. Greubel, Nov 11 2019
D-finite with recurrence: a(n) + 3*(-2*n+1)*a(n-1) = 0. - R. J. Mathar, Feb 24 2020
From Amiram Eldar, Jan 08 2022: (Start)
Sum_{n>=1} 1/a(n) = e^(1/6)*sqrt(3*Pi/2)*erf(1/sqrt(6)), where erf is the error function.
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)
MAPLE
seq(3^(n-1)*(2*n)!/(2^n*n!), n=1..20); # G. C. Greubel, Nov 11 2019
MATHEMATICA
Table[3^(n-1)*(2*n)!/(2^n*n!), {n, 20}] (* G. C. Greubel, Nov 11 2019 *)
PROG
(PARI) a(n) = prod(j=1, n, 6*j-3)/3; \\ Michel Marcus, Mar 13 2019
(Magma) F:=Factorial; [3^(n-1)*F(2*n)/(2^n*F(n)): n in [1..20]]; // G. C. Greubel, Nov 11 2019
(Sage) f=factorial; [3^(n-1)*f(2*n)/(2^n*f(n)) for n in (1..20)] # G. C. Greubel, Nov 11 2019
(GAP) F:=Factorial;; List([1..20], n-> 3^(n-1)*F(2*n)/(2^n*F(n))); # G. C. Greubel, Nov 11 2019
CROSSREFS
KEYWORD
easy,nonn
STATUS
approved