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%I #12 Oct 23 2022 16:10:54
%S 1,1,2,6,24,20,720,630,4480,36288,3628800,1925,479001600,444787200,
%T 387459072,5108103000,20922789888000,60988928000,6402373705728000,
%U 15205637551104,5516784599040000,2322315553259520000,1124000727777607680000,135277939046250
%N Divide n! by largest power of n which will leave the result an integer.
%F a(n) = n!/n^A011776(n) = A000142(n)/A060067(n).
%e a(12) = 1925 since 12! = 479001600 and dividing repeatedly by 12 gives 39916800, 3326400, 277200, 23100, 1925, 160.416666..., ...
%t Join[{1},Table[n!/n^IntegerExponent[n!,n],{n,2,30}]] (* _Harvey P. Dale_, May 01 2013 *)
%Y Cf. A000142, A011776, A060067.
%K nonn
%O 1,3
%A _Henry Bottomley_, Feb 19 2001
%E Offset corrected by _Sean A. Irvine_, Oct 23 2022