%I #12 Jun 24 2021 17:34:19
%S 1,1,2,3,12,20,120,315,1680,6048,60480,138600,1663200,9266400,
%T 69189120,340540200,5448643200,22870848000,411675264000,2111894104320,
%U 24135932620800,230388447744000,5068545850368000
%N a(n) = Product_{k=1..n} k/floor(n/k).
%C Sketch of proof that a(n) is an integer from Paul R. Pudaite, 9/28/2002: 1. n! = Product{p^([n/p]+[n/p^2]+...): prime p <= n}. 2. Product{[n/k]: k = 1...n} = Product{i^([n/i]-[n/i+1]): i=2...n}. 3. = Product{Product{Product{p^([n/i]-[n/i+1]): i such that p^k|i}: k such that p^k <= n}: prime p <= n}. 4. Reorganizing the exponents in the innermost product: ([n/p^k] - [n/(p^k+1)]) + ([n/(2 p^k)] - [n/(2 p^k + 1)] + ... = [n/p^k] - ([n/(p^k+1)] - [n/(2 p^k)]) - ... <= [n/p^k].
%F a(n) = n!/A010786(n).
%e a(6) = 6*5*4*3*2*1/([6/1]*[6/2]*[6/3]*[6/4]*[6/5]*[6/6]) = 6!/(6*3*2*1*1*1) = 20, where [x] denotes the greatest integer <= x.
%t Table[Product[k/Floor[n/k],{k,n}],{n,30}] (* _Harvey P. Dale_, Feb 27 2013 *)
%o (PARI) a(n) = prod(k=1, n, k/(n\k)); \\ _Michel Marcus_, Jun 24 2021
%Y Cf. A010786, A345684.
%K nonn
%O 1,3
%A _Clark Kimberling_, Sep 29 2002