asymptotic and monotonically increasing properties of prime factorization function?

Question

Questions

We define $A(x)= \text{number of prime factors of x}$

For example $A(2 \times 3^2) = 3$

I noticed when $s_k = \frac{N!}{\prod_j n_j}$ and $\sum_{j} n_j = N$

$$ s_1 < s_2 \implies A(s_1) < A(s_2)$$

Why is this so?

Also What is the asymptotic relation of $A(x)$ as $x \to \infty$?

My attempt for asymptotic relation

$S(x)=A(1)x + A(2)x^2 + A(3)x^3 + \dots$

$$G(x)=\frac{S(x)}{1-x} = A(1!)x + A(2!)x^2 + A(3!)x^3 + \dots$$

We note $A(n!)$ corresponds to the sequence: A022559

Using the result from there

$$ A(n!) \sim \ln\ln(n) + B_2 n + O(n)$$

Relevance

https://physics.stackexchange.com/questions/233634/number-theoretic-loophole-allows-alternative-definition-of-entropy

Answer 1

Counterexample: Let $N = 8$ and $s_1 = \binom83 = 56$, $s_2 = \binom84 = 70$, then $A(s_1) = 4 > 3 = A(s_2)$.

Answer 2

There obviously cannot be an asymptotic limit of your $A(n)$ as $n\to\infty$ because $A(n)=1$ whenever $n$ is prime and yet there are some $n$ like $2^k$ for which $A(n)$ can be arbitrarily large.

Nonetheless, if we look at Hardy and Wright's Introduction to the Theory of Numbers 5th ed. (1980), sections 22.10-22.11, we see that $A(n)$ is called there $\Omega(n)$. We are told that "the number of $n$, not exceeding $x$, for which $$|\Omega(n)-\log\log n|>(\log\log n)^{1/2+\delta}$$ is $o(x)$ for every positive $\delta$." In other words, $\log\log n$ is a good approximation to $\Omega(n)$ for most $n$.