Counting the number of pairs of $(i,j)$ such that $\operatorname{lcm}(i,j)=n$ Pairs $(i,j)$ such that $\operatorname{lcm}(i,j)=n$
how many pair $(i,j)$ can be formed such that $\operatorname{lcm}(i,j)=n$. here  $i\leq n$ and $j\leq n$.
here $n$ is a integer number less than $10^{14}$.

By way of adding commentary to  the answer by @Elaqqad I would like to
present a proof that indeed
$$\sum_{d|n} 2^{\omega(d)} = d(n^2).$$
This can be done using Dirichlet series and Euler products.
We have for $$2^{\omega(n)}$$ the Euler product
$$\prod_p 
\left(1 + \frac{2}{p^s} + \frac{2}{p^{2s}} + \frac{2}{p^{3s}}
+\cdots\right).$$
which is
$$\prod_p \left(-1 + 2\frac{1}{1-1/p^s}\right)
= \prod_p \frac{-1+1/p^s+2}{1-1/p^s}
\\ = \prod_p \frac{1+1/p^s}{1-1/p^s}
= \prod_p \frac{1-1/p^{2s}}{(1-1/p^s)^2}
= \frac{\zeta(s)^2}{\zeta(2s)}.$$
Therefore the Euler product for the LHS is
$$\frac{\zeta(s)^3}{\zeta(2s)}.$$
On the other hand, we get for $$d(n^2)$$ the Euler product
$$\prod_p 
\left(1 + \frac{3}{p^s} + \frac{5}{p^{2s}} + \frac{7}{p^{32}} 
+ \cdots\right)$$
which is
$$\prod_p \left(\sum_{q\ge 0} \frac{2q+1}{p^{qs}} \right)
= \prod_p \frac{1+1/p^{s}}{(1-1/p^{s})^2}
= \prod_p \frac{1-1/p^{2s}}{(1-1/p^{s})^3}
= \frac{\zeta(s)^3}{\zeta(2s)}$$
and we have  equality of the Dirichlet series  and their coefficients.
Since  $\omega(n) <\log_2 n$  and since  $d(n) \in  o(n^\epsilon)$ the
half-plane of convergence for these two is $\Re(s)> 2.$
Remark. If  we want to be  rigorous about it, we  have equality of
the coefficients  of the Dirichlet  series in the intersection  of the
two  half-planes  of  convergence  because for  the  Dirichlet  series
$\Lambda(s) = \sum_{n\ge 1} \frac{\lambda_n}{n^s}$ we have
$$\lambda_n = \frac{1}{2\pi i} 
\int_{c-i\infty}^{c+i\infty} \Lambda(s) n^{s-1} ds.$$
