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%I #7 Jul 11 2015 05:47:52
%S 2,144,2073600,17557585920000,192668014586363904000000,
%T 5116462645455544976110780416000000000,
%U 551096864092633744724294766310605805584384000000000000
%N Product of next n factorials.
%C Product of next n factorials gives this sequence, while product of first n factorials = superfactorial is A000178. If superfactorial is sf(n) then this function, "uppersuperfactorial", is usf(n)= sf(2n)/sf(n).
%F product(i!, i=n+1, 2n).
%F a(n) ~ 2^(2*n^2 + 5*n/2 + 5/12) * n^(n*(3*n/2+1)) * Pi^(n/2) / exp(n*(9*n+4)/4). - _Vaclav Kotesovec_, Jul 11 2015
%t Table[Product[i!, {i, n+1, 2n}], {n, 10}]
%Y Cf. A000178.
%K nonn
%O 1,1
%A _Zak Seidov_, Sep 22 2002
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