If $\displaystyle n_0=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ then $\displaystyle (m,n_0)\ne1 \Leftrightarrow p_1\mid m\text{ or }\cdots\text{ or }p_k\mid m$. Thus, if $\displaystyle A_{j,n}=\left\{m\leqslant n:p_j\mid m\right\}$ we see that

$\displaystyle [n]=E_n\sqcup\left(A_{1,n}\cup\cdots\cup A_{k,n}\right)$

where the square cup just means that $\displaystyle \left\{E_n,A_{1,n}\cup\cdots\cup A_{k,n}\right\}$ forms a partition of $\displaystyle [n]$. Thus, $\displaystyle a_n=n-\left|A_{1,n}\cup\cdots\cup A_{k,n}\right|$. But, by the inclusion exclusion principle

$\displaystyle \displaystyle \left|A_{1,n}\cup\cdots\cup A_{k,n}\right|=\sum_{j=1}^{k}(-1)^{j+1}\sum_{S\in S_j}\left|\bigcap_{s\in S}A_{s,n}\right|$

where $\displaystyle S_j=\left\{S\subseteq [k]:|S|=j\right\}$. Thus, noticing that any subcollection of $\displaystyle \{p_1,\cdots,p_k\}$ is a family of coprime numbers we see that

$\displaystyle \displaystyle \bigcap_{s\in S}A_{s,n}=\left\{m\leqslant n:p_s\mid m\text{ for all }s\in S\right\}=\left\{m\leqslant n:\prod_{s\in S}p_s\mid m\right\}$

From where simple reasoning leads us to

$\displaystyle \displaystyle\left|\bigcap_{s\in S}A_{s,n}\right|=\left\lfloor n\left(\prod_{s\in S}p_s\right)^{-1}\right\rfloor$

Thus,

$\displaystyle \displaystyle a_n=n-\sum_{j=1}^{k}(-1)^{j+1}\sum_{S\in S_j}\left\lfloor n\left(\prod_{s\in S}p_s\right)^{-1}\right\rfloor$

From where it quickly follows that

$\displaystyle \displaystyle \lim_{n\to\infty}\frac{a_n}{n}=1-\sum_{j=1}^{k}(-1)^{j+1}\sum_{S\in S_j}\left(\prod_{s\in S}p_s\right)^{-1}$

At least we know now that

$\displaystyle \displaystyle \prod_{j=1}^{k}\left(1-\frac{1}{p_j}\right)=1-\sum_{j=1}^{k}(-1)^{j+1}\sum_{S\in S_j}\left(\prod_{s\in S}p_s\right)^{-1}$

:)