%I #18 May 26 2024 08:29:43
%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(A004526(n)) = A055772(n)/A081123(n). [Corrected by _Amiram Eldar_, May 24 2024]
%e For n = 7, 7! = 5040 = 144*35, so 12 is its largest square-root-divisor, A000188(5040), and it is divisible by 6 = 3!, so a(7) = 12/3! = 2.
%t f[p_, e_] := p^Floor[e/2]; b[1] = 1; a[n_] := (Times @@ f @@@ FactorInteger[n!]) / Floor[n/2]!; Array[a, 100] (* _Amiram Eldar_, May 24 2024 *)
%Y Cf. A000142, A000188, A001405, A004526, A055772, A081123.
%K nonn
%O 1,6
%A _Labos Elemer_, Jul 25 2000