TITLE: $\lim_{n \to \infty}\frac{1}{n}\sum_{k = 1}^nf(k)$ where $f(n)$ is the largest prime factor exponent?
QUESTION [19 upvotes]: Let $f(n)$ the be largest exponent among exponents of the prime factor of $n$. E.g. $f(80) = 4$ since $80 = 2^4.5$ and the prime factor of $80$ which has the largest exponent is $2$ which occurs with the exponent $4$. Trivially for all square-free numbers $n, f(n) = 1$. Experimental data for $n \le 3.5 \times 10^9$ shows that.
$$
\mu = \lim_{n \to \infty}\frac{1}{n}\sum_{k = 1}^nf(n) \approx 1.784744
$$
Q 1: Does this limit exist and does it have a closed form?
If all positive integers were square-free, the above mean would have been exactly $1$. But this is not the case because there are non-squarefree numbers. Thus, we can say that the square-free numbers contribute exactly $1$ to the above mean value while all numbers which have prime factors with an exponent $> 1$ contribute the remaining $0.784744$.
Using the natural density of $k$-th power free numbers, I can show that
$$
\mu \ge \sum_{k = 1}^{\infty}k \Big(\frac{1}{\zeta(k+1)} - \frac{1}{\zeta(k)}\Big) \approx 1.70521
$$
Q 2: Is it possible to show a weaker result like $\mu < 2$?

REPLY [8 votes]: There is an elementary solution.
$$\sum_{n=1,f(n)\le k}^\infty n^{-s} = \prod_p (1+\sum_{d=1}^k p^{-sd}) = \frac{\zeta(s)}{\zeta((k+1)s)}$$
$$\sum_{n=1}^\infty f(n) n^{-s}= \sum_{k=0}^\infty \sum_{n=1,f(n)> k}^\infty n^{-s}
= \sum_{k=0}^\infty
(\zeta(s)-\frac{\zeta(s)}{\zeta((k+1)s)})\\=-1+\zeta(s) \sum_{n=1}^\infty a(m)m^{-s}$$
Where

$\sum_{n=1}^\infty a(m)m^{-s}=1+\sum_{k=1}^\infty
(1-\frac{1}{\zeta((k+1)s)})$ converges absolutely for $\Re(s) > 1/2$

$ \sum_{m\le x} |a(m)|=o(x)$

$\sum_{n=1}^\infty \frac{a(m)}{m}\ne 0$


Whence
$$\sum_{n\le x} f(n) = -1+ \sum_{m \le x} a(m) \lfloor x/m\rfloor= 
x\sum_{m \le x} \frac{a(m)}m+O(1+\sum_{m \le x} |a_m|)\\= x \sum_{m=1}^\infty \frac{a(m)}{m}\ +o(x)$$