$\text{Notations}$
Let $\pi(n)$ be the prime countiong function.
Let $\alpha(n)$ denote the number of prime factors of $n$ and $\beta(n)$ the sum of the prime factors of $n$. In other words, if $$n=p_1^{x_1}p_2^{x_2}...p_m^{x_m}$$ then $\alpha(n)=m$ and $\beta(n)=p_1+p_2+...+p_m$
(I changed the notations; It was pointed out in the comments that $\psi$ and $\omega$ are other functions and they were misleading)
$\text{Statement}$
Let $\pi(n)$ be the prime counting function, $\psi(n)$ the number of prime factors and $\omega(n)$ the sum of the prime factors of $n$. Prove or disprove that there are infinitely many $n$ such that: $$\pi(n)=\alpha(n)\beta(n)$$
$\text{Some other observations}$
Here are the first solutions:
4, 120, 437, 546, 620, 8144, 11509, 170049, 170907, 340655, 478476, 789575
Also, if we use the Prime Number Theorem ($\pi(n)$ is aproximatively $\frac{n}{\ln(n)}$) we can derive solutions in an exhaustive manner. For example, find at what point $\frac{n}{\pi(n)}$ is aproximatively $\frac{47}{2}$ and then subtract $47$ from $n$ until you get the solution. Using this method we get $45764089927$ which in fact works.
$\text{Why is it an interesting problem?}$
I find it rather interesting due to the unusual relationship between these $3$ functions. I played a while with several arithmetic functions and couldn't find strong conection between them, but the equation $\pi(n)=\alpha(n)\cdot\beta(n)$ seems promising.
Here is a naive start I came up with (which is nothing but crude bounding) but it shows that if we divide such an $n$ by the largest prime divisor of itself, the remaining part should be small enough.
Suppose $p_m$ is the largest prime. Note that $\omega(n) \le m p_m$, so $\pi(n) \le m^2 p_m$, and $\omega(n) \ge p_m$, so $\pi(n) \ge m p_m \ge p_m$. Now note that $n \ge p_1 \cdots p_m \ge 2^m$, so $m \le \log_2 n$. Since for large enough $n$, $\pi(n)$ is bounded between $(1 \pm \varepsilon) \cdot \frac{n}{\ln n}$, we have the bounds $\frac{\ln n}{1 + \varepsilon} \le \frac{n}{p_m} \le \frac{(\ln n) \cdot (\log_2 n)^2}{1 - \varepsilon}$, and as a consequence of this, the fraction mentioned should be bounded above by the order of $(\ln n)^3$ and below by the order of $\ln n$.