Product of greatest common divisors As usually, let $\gcd(a,b)$ be the greatest common divisor of integer numbers $a$ and $b$.
What is the asymptotics of
$$\left(\prod_{i=1}^{i=n} \prod_{j=1}^{j=n} \gcd(i,j)\right)^{1/n^2}
$$
as $n \to \infty?$
 A: This is a pretty problem and I would like to contribute an answer that differs from the older posts and let those StackExchange wizards evaluate its correctness.
First, work with the logarithm $S$ of the product $P$, which is
$$ S = \log P = 
\frac{1}{n^2} 
\left( -\log n! + 2 \sum_{1\le i\le j \le n} \log\gcd(i, j)\right).$$
Now introduce the function $p_k(n)$ which gives the number of integers $q$ between $1$ and $n$ that have $\gcd(q,n) = k,$ i.e.
$$p_k(n) = \#\left\{q\;|\; \gcd(q,n)=k \; \wedge \; 1\le q\le n\right\}.$$
It has the following useful property:
$$\sum_{d|n} p_k(d) =
\begin{cases}
0 \quad\text{if}\quad k\nmid n \\
n/k \quad\text{otherwise}.
\end{cases}.$$
To see this, do the following evaluation:
$$ \sum_{d|n} p_k(d) =  \sum_{d|n} \sum_{q=1 \atop \gcd(q, d)=k}^d 1 =
\sum_{f|n/k} \sum_{q=1\atop \gcd(q, kf)=k}^{kf} 1 \\=
\sum_{f|n/k} \sum_{r=1\atop \gcd(kr, kf)=k}^{f} 1 =
\sum_{f|n/k} \sum_{r=1\atop \gcd(r, f)=1}^{f} 1 = 
\sum_{f|n/k} \phi(f) = n/k.$$
It follows that the Dirichlet series $L_k(s)$ for $p_k(n)$ is given by
$$ L_k(s) = \frac{1}{\zeta(s)} \sum_{m\ge 1} \frac{m}{(mk)^s} = 
\frac{1}{k^s} \frac{\zeta(s-1)}{\zeta(s)}.$$
The sum term in $S$ is equal to
$$\sum_{k=1}^n \log k \times \sum_{m=1}^n p_k(m) =
 \sum_{m=1}^n \sum_{k=1}^n  \log k \times p_k(m) =
 \sum_{m=1}^n \sum_{k=1}^m  \log k \times p_k(m)$$
The last equality holds because $m\le n$ and we require $k\le m$, so that we may replace $n$ in the upper limit by $m.$
But we have $$\sum_{k=1}^\infty \log k \times L_k(s) =
-\zeta'(s) \frac{\zeta(s-1)}{\zeta(s)}.$$
In the above the term generated as the coefficient of $n^s$ is
$$\sum_{k=1}^\infty \log k \times p_k(n) = \sum_{k=1}^n \log k \times p_k(n),$$
which is precisely what we need, since $p_k(n)=0$ when $k> n,$ so that in fact the sum only goes up to $n$ as required.
This means that we may apply the Wiener-Ikehara theorem to the sequence
$$a_m = \sum_{k=1}^m \log k \times p_k(m),$$
getting $$\sum_{m=1}^n a_m \sim 
\operatorname{Res}\left(-\zeta'(s) \frac{\zeta(s-1)}{\zeta(s)}; s=2\right) \frac{n^2}{2} =
- \frac{6\zeta'(2)}{\pi^2}  \frac{n^2}{2}.$$
We see that this dominates the term in $\log n!$ and we obtain in the limit that
$$S = \log P = - \frac{6\zeta'(2)}{\pi^2} \approx 0.5699609929$$
and hence $$P = \exp\left(- \frac{6\zeta'(2)}{\pi^2}\right) \approx 1.768198078.$$
Very pretty indeed.
A: You can show that your limit approaches 
$$\prod_{p} p^{1/(p^2-1)}$$
(where $p$ ranges over all positive primes). I'm not sure if this has a closed form, but evaluating this for $p$ up to $10^5$ gives a value of $1.78\dots$ for this constant, so I don't believe the value $5/2\sqrt{2}$ is correct. 
To show that your limit approaches my product above, pick a prime $p$, and consider when $p | \gcd(i, j)$. Note that this happens exactly when both $i$ and $j$ are divisible by $p$, which happens for $\lfloor n/p \rfloor^2$ pairs $(i, j)$ with $1 \leq i, j \leq n$. Similarly, $p^2 | \gcd(i, j)$ when both $i$ and $j$ are divisible by $p^2$, which happens for $\lfloor n/p^2 \rfloor^2$ pairs $(i,j)$, and in general $p^k | \gcd(i, j)$ for $\lfloor n/p^k \rfloor^2$ pairs $(i, j)$. It follows that the exponent of $p$ in the prime factorization of your product is equal to
$$\sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k}\right\rfloor^2$$
Taking the logarithm of your expression, it therefore becomes
$$\dfrac{1}{n^2}\sum_{p}\left(\sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k}\right\rfloor^2\right)\log p$$
Now, note that
$$\sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k}\right\rfloor^2 = O(n\log n) + \sum_{k=1}^{\infty} \frac{n^2}{p^{2k}} = O(n\log n) + \dfrac{n^2}{p^2 - 1}$$
Substituting this back in, we find that as $n\rightarrow \infty$, the logarithm of your expression will tend to
$$\sum_{p} \frac{\log p}{p^2 - 1}$$
and therefore the expression itself will tend to
$$\prod_{p} p^{1/(p^2-1)}$$
A: Just looking numerically (no reasoning), it seems to converge, to somewhere around $1.76$:



