Some regularity in the prime decomposition

It's quite normal that the summatory function of an irregularly behaving function is rather regular, since the summation cancels deviations from the average behaviour. The irregularity of $\rho$ is rather benign, since for every integer $m \geqslant 1$, we get a straight line

$$\rho(n) = \rho(m) + \frac{n}{m}$$

from the numbers $n = m\cdot p$, where $p$ is a prime not dividing $m$. Since almost all numbers have few prime factors, only a negligible fraction of $n \leqslant x$ have $\rho(n)$ not lying on one of the first few lines (where "first few" of course depends on $x$).

To determine the asymptotic behaviour of $\Phi$, it is useful to write

$$\Phi(x) = \sum_{p \leqslant x} F(p,x)\cdot p,$$

where the summation is over the primes not exceeding $x$, and $F(p,x)$ counts how often $p$ occurs in the sum defining $\Phi$. Considering a fixed prime $p$, we note that $p$ occurs in the summand $\rho(n)$ if and only if $p\mid n$. There are $\bigl\lfloor \frac{x}{p}\bigr\rfloor$ multiples of $p$ not exceeding $x$. $p$ occurs at least twice in $\rho(n)$ if and only if $p^2 \mid n$. And so on, $p$ occurs at least $k$ times in $\rho(n)$ if and only if $p^k\mid n$. That may be familiar from counting the number of times $p$ occurs in the factorisation of $\lfloor x\rfloor!$, and indeed $F(p,x)$ is that same number, i.e.

$$F(p,x) = \sum_{k = 1}^{\infty} \biggl\lfloor \frac{x}{p^k}\biggr\rfloor.$$

The sum contains only finitely many nonzero terms, of course.

We get an upper bound $F(p,x) < \frac{x}{p-1}$ from just ignoring the $\lfloor\,\cdot\,\rfloor$, and thus an upper bound

$$\Phi(x) \leqslant \sum_{p \leqslant x} \frac{x}{p-1}\cdot p = x\cdot \pi(x) + x\sum_{p \leqslant x} \frac{1}{p-1}.\tag{1}$$

By Mertens' theorem,

$$\sum_{p \leqslant x} \frac{1}{p-1} = \sum_{p \leqslant x} \frac{1}{p} + \sum_{p \leqslant x} \frac{1}{p(p-1)} = \log \log x + O(1),$$

so the second term on the right of $(1)$ is negligible compared to the error made by approximating $\pi(x)$ by $\frac{x}{\log x}$ or $\operatorname{Li}(x)$. We have $\Phi(x) \in O\bigl(\frac{x^2}{\log x}\bigr)$, and the lower bound given by Ahmad shows $\Phi(x) \in \Theta\bigl(\frac{x^2}{\log x}\bigr)$, and we only have an uncertainty factor of $2$ for the coefficient.

We note that

$$\Phi(x) - \sum_{p \leqslant x} \biggl\lfloor \frac{x}{p}\biggr\rfloor \cdot p$$

is negligible compared to the leading term, since

$$\sum_{p \leqslant x} \sum_{k = 2}^{\infty} \biggl\lfloor \frac{x}{p^k}\biggr\rfloor\cdot p \leqslant \sum_{p \leqslant x} \frac{x}{p-1} \in O(x\log \log x).$$

For a more precise analysis of the dominant term, let

$$S(y) := \sum_{p \leqslant y} p.$$

From the prime number theorem with error bounds, one finds $S(y) = \operatorname{Li}(y^2) + O\bigl(y^2e^{-c\sqrt{\log y}}\bigr)$ with some $c > 0$ via summation by parts. Another summation by parts gives

\begin{align} \sum_{p \leqslant x} \biggl\lfloor \frac{x}{p}\biggr\rfloor \cdot p &= \sum_{m \leqslant x} \sum_{\frac{x}{m+1} < p \leqslant \frac{x}{m}} m\cdot p \\ &= \sum_{m \leqslant x} m\biggl(S\biggl(\frac{x}{m}\biggr) - S\biggl(\frac{x}{m+1}\biggr)\biggr) \\ &= \sum_{m \leqslant x} S\biggl(\frac{x}{m}\biggr). \end{align}

Let $\alpha \in (0,1)$. The trivial estimate $S(y) < y^2$ for all $y \geqslant 1$ gives

$$\sum_{x^{\alpha} < m \leqslant x} S\biggl(\frac{x}{m}\biggr) < x^2 \sum_{m > x^{\alpha}} \frac{1}{m^2} \sim x^{2-\alpha},$$

so these terms don't contribute to the main asymptotic behaviour. Neither does the sum of the $O\bigl(y^2 e^{-c\sqrt{\log y}}\bigr)$ error terms, since $\log (x/m) \geqslant (1-\alpha)\log x$ for $m \leqslant x^{\alpha}$, so

$$\sum_{m \leqslant x^{\alpha}} O\biggl(\frac{x^2}{m^2} e^{-c\sqrt{\log (x/m)}}\biggr) \leqslant K x^2 e^{-c\sqrt{1-\alpha}\sqrt{\log x}}\sum_{m \leqslant x^{\alpha}} \frac{1}{m^2} \in O\bigl(x^2e^{-d\sqrt{\log x}}\bigr)$$

with $d = c\sqrt{1-\alpha} > 0$. Thus we need to look at

$$\sum_{m \leqslant x^{\alpha}} \operatorname{Li}\biggl(\frac{x^2}{m^2}\biggr).$$

With $\operatorname{Li}(y) = \frac{y}{\log y} + O\bigl(\frac{y}{(\log y)^2}\bigr)$, we see

$$\sum_{m \leqslant x^{\alpha}} \operatorname{Li}\biggl(\frac{x^2}{m^2}\biggr) = \sum_{m \leqslant x^{\alpha}} \frac{x^2}{2m^2\log (x/m)} + O\biggl(\frac{x^2}{(\log x)^2}\biggr),$$

and

$$\sum_{m \leqslant x^{\alpha}} \frac{x^2}{2m^2\log (x/m)} \sim \sum_{m} \frac{x^2}{2m^2\log x} = \frac{\pi^2}{12} \cdot \frac{x^2}{\log x}\tag{$\ast$}$$

then shows

$$\Phi(x) \sim \frac{\pi^2}{12}\cdot \frac{x^2}{\log x}.\tag{2}$$

To see the asymptotic equivalence in $(\ast)$, split the sum at $m \approx \log x$.