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%I #6 Dec 12 2021 19:45:33
%S 1,1,1,1,1,2,2,1,3,6,6,2,2,2,6,3,3,2,2,2,2,2,2,2,10,10,30,30,30,12,12,
%T 3,3,6,30,10,10,10,30,6,6,2,2,2,30,60,60,30,210,42,42,42,42,28,28,2,2,
%U 4,4,4,4,4,84,21,21,14,14,14,42,6,6,2,2,2,10,10,70,140,140,14,126,126
%N Let k be largest number such that k^2 divides n!; a(n) = k/floor(n/2)!.
%F a(n) = A000188(n!)/floor(n/2)! = A055772(n)/(A000142(A001057(n))^2.
%e n=7, 7! = 5040 = 144*35, so 12 is its largest square-root-divisor, A000188(5040) and it is divisible by 6=3! and a(7)=2, the quotient, 12/3!.
%Y Cf. A000142, A000188, A001057, A055772, A001405.
%K nonn
%O 1,6
%A _Labos Elemer_, Jul 25 2000
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