How to compute sum of sum of gcd of factor pairs of a number upto a large number efficiently?

It can be proved that $displaystyle sum_{dmid n} (d,n/d) = sum_{d^2mid n}tau(n/d^2)varphi(d)$.

So one can show that $displaystyle F(x) = sum_{nle sqrt{x}}varphi(n)sum_{m le x/n^2} leftlfloorfrac{x}{n^2 m}rightrfloor$.

This is the best i can do, but this is still computationally unfeasible for $x approx 10^{15}$.

PARTIAL ANSWER: Here is an alternative formula for $F(x)$: $$F(x)=sum_{k=1}^{sqrt{x}}g_x(k)k$$ where $$g_x(k) = |{ (a,b) : abk^2 le x, gcd(a,b)=1 }|$$

proof:

For a fixed $x>0$, consider the following set $$I_x={ (k, d, n) : k=gcd(d, n/d), d|n, n le x}$$ Then your $F(x)$ is just $$F(x)= sum_{(k,d,n) in I_x} k$$ Let's study how this set $I_x$ is made.

First of all, note that for all $(k,d,n) in I_x$ you have that $k$ divides both $d$ and $n/d$, hence $$k^2 mbox{ divides } n = d cdot (n/d)$$ In particular $k le sqrt{x}$.

On the other hand, for arbitrary $k le sqrt{x}$ you have $(k,k,k^2) in I_x$. This means that all numbers $k le sqrt{x}$ appear at least once as the first coordinate of a triple $(k,d,n) in I_x$, while all numbers $k > sqrt{x}$ don't.

So, let's call $$g_x(k) = |{ (d,n) : (k,d,n) in I_x }|$$ This function counts how many times $k$ appears as a first coordinate of a triple $(k,d,n) in I_x$, so that $$F(x)= sum_{(k,d,n) in I_x} k=sum_{k=1}^{sqrt{x}} g_x(k) cdot k$$ To conclude the proof we have to show that $$g_x(k) = 2 lfloor frac{x}{k^2} rfloor-1$$

For a fixed $k le sqrt{x}$, you have that $(k,d,n) in I_x$ if and only if $k= gcd(d,n/d)$. This means that $d=ak$ and $n/d=bk$ for some $a,b$. Thus we can consider the set of quintuples $$J_x= { (k,a,b,d,n) : d=ak, n/d=bk, gcd(a,b)=1, d|n, n le x }$$ which is in clear bijection with $I_x$ by the map $(k,a,b,d,n) mapsto (k,d,n)$. Note that $a=d/k$ and $b=n/(dk)=n/(abk^2)$. So that our $J_x$ is in bijection with the set $$L_x = { (k, a, b) : abk^2 le x , gcd(a,b)=1}$$ by the map $(k,a,b,d,n) mapsto (k,a,b)$, because $n=abk^2 le x$. In other words $g_x(k)$ counts the number of pairs $(a,b)$ of coprime numbers $a,b$ such that $abk^2 le x$, or $$ab le frac{x}{k^2}$$

continues...

OK, MY BAD, NOW I NOTICED THAT THIS NUMBER IS NOT $2 lfloor frac{x}{k^2} rfloor-1$, BUT IT'S TRICKIER. I'll leave this answer for who wants to conclude my computations.