The limit of $\frac{|A_n|}{n^2}$

If $n=1$, then $|A_n|=3$. From now on $n>1$. Denote $$ A_n^{<,j}=\{(i,j)\in A_n: i<j\}\\ A_n^{=}=\{(i,j)\in A_n: i=j\}\\ A_n^{>,i}=\{(i,j)\in A_n: i>j\} $$ Then from definition of Euler's function it follows that $|A_n^{<,j}|=\varphi(j)$ and $|A_n^{>,i}|=\varphi(i)$. Obviously $|A_n^{=}|=1$, so $$ |A_n|=|A_n^{=}|+\sum\limits_{j=1}^n|A_n^{<,j}|+\sum\limits_{i=1}^n|A_n^{>,i}|=1+2\sum\limits_{k=1}^n\varphi(k) $$ Then the desired limit is equal to $$ \lim\limits_{n\to\infty}\frac{|A_n|}{n^2}=2\lim\limits_{n\to\infty}\frac{1}{n^2}\sum\limits_{k=1}^n\varphi(k) $$ As wikipedia says, $$ \sum\limits_{k=1}^n\varphi(k)=\frac{3n^2}{\pi^2}+\mathcal{O}\left(n(\log n)^{2/3}(\log\log n)^{4/3}\right) $$ hence the answer is $6\pi^{-2}$