GCD(k,j) Product

[Leroy Quet]

>Analogous to sequence A018806, but with the double sum replaced with a 
>double product,
>
>we can have the sequence:
>
>a(n) =
>
>
>product{k=1 to n} product{j=1 to n}  GCD(k,j),
>
>which begins, I believe:
>
>1, 2, 6, 96, 480,...,
>
>which is not in the EIS.
>
>
>Am I right to assume that:
>
>limit{n-> oo}  a(n)^(1/n^2) =
>
>exp(-zeta'(2)/zeta(2)) ?
>
>(where zeta(2) = pi^2/6,
>and zeta'(2) is the derivative of the Riemann zeta function at 2)
>
>


PS: a(n) also is:

product{k=1 to n} product{j=1 to n} 
                         LCM(1,2,3,...,floor(min(n/k,n/j))),

I believe.


And, if p^r = highest power of p (p=prime) which divides a(n), then:

r = sum{k=1 to floor(ln(n)/ln(p))} (floor(n/p^k))^2,
 
I also believe.

thanks,
Leroy Quet