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a(n) = Product_{i=1..n} floor(3*i/2).
3

%I #17 Oct 03 2018 05:41:04

%S 1,3,12,72,504,4536,45360,544320,7076160,106142400,1698278400,

%T 30569011200,580811212800,12197035468800,268334780313600,

%U 6440034727526400,161000868188160000,4347023441080320000,121716656350248960000,3651499690507468800000

%N a(n) = Product_{i=1..n} floor(3*i/2).

%F a(n) ~ (3/2)^n * n! * 2^(1/6) * sqrt(Pi) / (Gamma(1/3) * n^(1/6)).

%F Recurrence: 4*a(n) - 6*a(n-1) - 3*(n - 1)*(3*n - 4)*a(n-2) = 0, with n >= 3. - _Bruno Berselli_, Oct 03 2018

%t Table[Product[Floor[i*3/2], {i, 1, n}], {n, 1, 20}]

%t RecurrenceTable[{4 a[n] - 6 a[n - 1] - 3 (n - 1) (3 n - 4) a[n - 2] == 0, a[1] == 1, a[2] == 3}, a, {n, 1, 20}] (* _Bruno Berselli_, Oct 03 2018 *)

%Y Cf. A010786, A180736, A275062, A319949, A319950, A317980.

%K nonn

%O 1,2

%A _Vaclav Kotesovec_, Oct 02 2018