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A224497 a(n) = sqrt(floor(n/2)! * Product_{k=1..n} Product_{i=1..k-1} gcd(k,i)).

%I A224497 #12 Aug 02 2019 05:27:14
%S A224497 1,1,1,1,2,2,12,12,96,288,5760,5760,829440,829440,46448640,2090188800,
%T A224497 267544166400,267544166400,346737239654400,346737239654400,
%U A224497 1109559166894080000,209706682542981120000,73816752255129354240000,73816752255129354240000
%N A224497 a(n) = sqrt(floor(n/2)! * Product_{k=1..n} Product_{i=1..k-1} gcd(k,i)).
%C A224497 The order of the primes in the prime factorization of a(n) is given by
%C A224497 ord_{p}(a(n)) = (1/4)*Sum_{i>=1} floor(n/p^i)*(floor(n/p^i)-1) + (1/2)*Sum_{i>=1} floor(floor(n/2)/p^i).
%C A224497 For n > 1: a(n) = a(n-1) if and only if n is prime.
%F A224497 a(n) = sqrt(floor(n/2)! * A224479(n)).
%F A224497 A092287(n) = A056040(n) * a(n)^4.
%p A224497 A224497 := n -> sqrt(iquo(n,2)!*mul(mul(igcd(k,i), i=1..k-1), k=1..n)):
%p A224497 seq(A224497(i), i = 0..23);
%o A224497 (Sage)
%o A224497 def A224497(n):
%o A224497     R = 1;
%o A224497     for p in primes(n):
%o A224497         s = 0; t = 0
%o A224497         r = n; u = n//2
%o A224497         while r > 0 :
%o A224497             r = r//p; u = u//p
%o A224497             t += u; s += r*(r-1)
%o A224497         R *= p^((t+s/2)/2)
%o A224497     return R
%o A224497 [A224497(i) for i in (0..23)]
%Y A224497 Cf. A224479.
%K A224497 nonn
%O A224497 0,5
%A A224497 _Peter Luschny_, Apr 08 2013