The number of primes in the factorization of $N!$ Is there an approximation to the number of primes in the factorization of $N!$?
For example:


*

*For $N=10$, this number is $15$.

*For $N=100$, this number is $239$.

*For $N=1000$, this number is $2877$.

*For $N=10000$, this number is $31985$.

*For $N=100000$, this number is $343614$.

*For $N=1000000$, this number is $3626619$.


There seem to be a gradual ascendant towards $4N$, but has that been proved as an upper limit?

This does not completely answer your question but is helpful to know and the argument can be made completely rigorous.
The highest power of a prime $p$ dividing $N!$ is
$$\left\lfloor \dfrac{N}p \right\rfloor + \left\lfloor \dfrac{N}{p^2} \right\rfloor + \cdots \sim \dfrac{N}{p-1}$$ and the number of primes less than $N$ is $\sim \dfrac{N}{\log(N)}$ and $\displaystyle \sum_{p \leq N} \dfrac1p \sim \ln(\ln(N))$
and hence
$$\sum_{p \leq N} \dfrac{N}{p-1} \sim N \ln(\ln(N))$$
Hence, my possible guess is $\sim N \ln(\ln(N))$.

For instance, if you take $N$ to be $10^m$, we then have the highest power of $2$ dividing $10^m$ as $$\dfrac{N}2 + \dfrac{N}{2^2} + \cdots \dfrac{N}{2^m} = \dfrac{N}2 \dfrac{1-1/2^m}{1-1/2} = N - \dfrac{N}{2^m}$$
Similarly, the highest power of $5$ dividing $10^m$ as $$\dfrac{N}5 + \dfrac{N}{5^2} + \cdots \dfrac{N}{5^m} = \dfrac{N}5 \dfrac{1-1/5^m}{1-1/5} = \dfrac{N}4 - \dfrac{N}{4 \cdot 5^m}$$
Hence, the number of prime factors of $N = 10^m$ is $$\dfrac{5N}4 - 5^m - 2^{m-2}$$
