We define the arithmetic function: $\displaystyle f(n) = \max\limits_{p^{\alpha} || n} \alpha$, that is if $\displaystyle n = p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ (prime factorization of $n$) then $f(n) = \max\limits_{1 \le i \le k} \alpha _i$.
How do we determine the asymptotic growth of the summatory function of $f$, that is $$\displaystyle S(x) = \sum\limits_{1 \le n \le x} f(n)$$
Is it possible to express the Dirichlet series corresponding to $f$, that is $\displaystyle \sum\limits_{1 \le n} \frac{f(n)}{n^s}$ in terms of Dirichlet series of some known arithmetic functions ?
If we group the numbers by value of $f(n)$, we have
$$S(x) = \sum_{k \leqslant \log_2 x} k\cdot\operatorname{card} \{ n \leqslant x : f(n) = k\} = \sum_{k \leqslant \log_2 x} \operatorname{card} \{ n \leqslant x : f(n) \geqslant k\}.$$
If $p(k,x)$ denotes the count of numbers $\leqslant x$ that are multiples of a $k$-th power of a prime, that becomes
$$S(x) = \sum_{k} p(k,x).$$
Since
$$\lfloor x\rfloor - p(k,x) = \sum_{ n\leqslant x} \mu(n)\left\lfloor\frac{x}{n^k} \right\rfloor$$
for $k\geqslant 2$, we have, omitting the upper bound on the sums, since all later terms are $0$,
\begin{align} S(x) &= \sum_{k=1}^\infty p(k,x)\\ &= \lfloor x\rfloor - 1 + \sum_{k=2}^\infty p(k,x)\\ &= \lfloor x\rfloor - 1 - \sum_{k=2}^\infty \left(\sum_{m=2}^\infty \mu(m)\left\lfloor \frac{x}{m^k}\right\rfloor\right)\\ &= \lfloor x\rfloor -1 - \sum_{k=2}^\infty \left(\sum_{m=2}^\infty \mu(m)\frac{x}{m^k} - \sum_{m=2}^\infty \mu(m)\biggl(\frac{x}{m^k} - \left\lfloor\frac{x}{m^k}\right\rfloor\biggr)\right)\\ &= x\biggl(1 - \sum_{k=2}^\infty \sum_{m=2}^\infty \frac{\mu(m)}{m^k}\biggr) - (1+x-\lfloor x\rfloor) + \sum_{k=2}^\infty \sum_{m=2}^\infty \mu(m)\biggl(\frac{x}{m^k} - \left\lfloor \frac{x}{m^k}\right\rfloor\biggr)\\ &= \Biggl(\sum_{k=1}^\infty \biggl(1 - \frac{1}{\zeta(k)}\biggr)\Biggr)\cdot x + O(1) + O(\sqrt{x}). \tag{1} \end{align}
It remains to prove the $O(\sqrt{x})$ estimate for the last sum. We have
\begin{align} \left\lvert \sum_{k=2}^\infty \sum_{m=2}^\infty \mu(m)\biggl(\frac{x}{m^k} - \left\lfloor \frac{x}{m^k}\right\rfloor\biggr)\right\rvert &\leqslant \sum_{k=2}^\infty \sum_{m=2}^\infty \biggl(\frac{x}{m^k} - \left\lfloor\frac{x}{m^k}\right\rfloor\biggr)\\ &= \sum_{m=2}^\infty \sum_{k=2}^\infty \biggl(\frac{x}{m^k} - \left\lfloor\frac{x}{m^k}\right\rfloor\biggr)\\ &= \sum_{m=2}^{\lfloor \sqrt{x}\rfloor}\sum_{k=2}^\infty \biggl(\frac{x}{m^k} - \left\lfloor\frac{x}{m^k}\right\rfloor\biggr) + \sum_{m=\lfloor\sqrt{x}\rfloor+1}^\infty\sum_{k=2}^\infty \frac{x}{m^k}. \end{align}
The last sum can be explicitly evaluated,
$$\sum_{m= a+1}^\infty \sum_{k=2}^\infty \frac{1}{m^k} = \sum_{m=a+1}^\infty \frac{1}{m(m-1)} = \frac{1}{a},$$
and so
$$\sum_{m=\lfloor \sqrt{x}\rfloor+1}^\infty\sum_{k=2}^\infty \frac{x}{m^k} = \frac{x}{\lfloor\sqrt{x}\rfloor} \in O(\sqrt{x}).$$
For the other sum, we split the inner sum at $\frac{\log x}{\log m}$. For $k \leqslant \frac{\log x}{\log m}$, we use $0 \leqslant z-\lfloor z\rfloor < 1$ to obtain the estimate
$$\sum_{m=2}^{\lfloor \sqrt{x}\rfloor} \frac{\log x}{\log m},$$
which with
$$\sum_{m=2}^a \frac{1}{\log m} = \frac{a}{\log a} + O\left(\frac{a}{(\log a)^2}\right)$$
yields a bound
$$\sum_{m=2}^{\lfloor\sqrt{x}\rfloor}\sum_{k=2}^{\left\lfloor\frac{\log x}{\log m}\right\rfloor}\biggl(\frac{x}{m^k}-\left\lfloor\frac{x}{m^k}\right\rfloor\biggr) \leqslant 2\sqrt{x} + O\left(\frac{\sqrt{x}}{\log x}\right)\in O(\sqrt{x}),$$
and for the remaining part we use
$$\sum_{k=a}^\infty \frac{1}{m^k} = \frac{1}{m^a(1-m^{-1})}$$
to obtain
$$\sum_{m=2}^{\lfloor\sqrt{x}\rfloor}\sum_{k=\left\lfloor\frac{\log x}{\log m}\right\rfloor+1}^\infty \frac{x}{m^k} \leqslant \sum_{m=2}^{\lfloor\sqrt{x}\rfloor} \frac{x}{m^{\left\lfloor\frac{\log x}{\log m}\right\rfloor+1}(1-m^{-1})}\leqslant \sum_{m=2}^{\lfloor\sqrt{x}\rfloor} \frac{1}{1-m^{-1}} \leqslant 2\lfloor\sqrt{x}\rfloor.$$
Thus the asymptotic $(1)$ is proved.
For the Dirichlet series, we can - for $\operatorname{Re} s > 1$ - write
$$\sum_{n=1}^\infty \frac{f(n)}{n^s} = \sum_{k=1}^\infty \underbrace{\sum_{n=1}^\infty\frac{[f(n) \geqslant k]}{n^s}}_{\sigma_k(s)},$$
where $[\,\cdot\,]$ is the Iverson bracket. Since $[f(n)\geqslant 1] = 1$ for all $n\geqslant 2$ and $f(1) = 0$, we have $\sigma_1(s) = \zeta(s) - 1$, and for $k\geqslant 2$ we note that
\begin{align} \sigma_k(s) &= \sum_{m=2}^\infty (-\mu(m))\sum_{n=1}^\infty \frac{1}{(m^kn)^s}\\ &= \sum_{m=2}^\infty \frac{-\mu(m)}{m^{ks}}\zeta(s)\\ &= \biggl(1 - \sum_{m=1}^\infty \frac{\mu(m)}{m^{ks}}\biggr)\zeta(s)\\ &= \biggl(1 - \frac{1}{\zeta(ks)}\biggr)\zeta(s), \end{align}
hence
$$\sum_{n=1}^\infty \frac{f(n)}{n^s} = \zeta(s)\cdot\sum_{k=1}^\infty\Bigl(1-\frac{1}{\zeta(ks)}\Bigr) = \zeta(s)\cdot \sum_{k=1}^\infty \frac{\zeta(ks)-1}{\zeta(ks)}.$$