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A056043 Let k be largest number such that k^2 divides n!; a(n) = k/[n/2]!. 0

%I

%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/[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 A000142, A000188, A001057, A055772, A001405.

%K nonn

%O 1,6

%A _Labos Elemer_, Jul 25 2000

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Last modified August 6 09:02 EDT 2020. Contains 336228 sequences. (Running on oeis4.)