prime numbers – Oppermann’s conjecture in an imaginary world?

I apologize for such a broad title. I just didn’t know how to fit a brief specific title for the following:

For $𝑥∈𝑁$ where $𝑁$ are all the natural numbers.

The Oppermann’s conjecture states that, for every integer x > 1, there is at least one prime number between $x(x − 1)$ and $x^2$, and at least another prime between $x^2$ and $x(x + 1)$.

The difference between $x(x − 1)$ and $x^2$ is $x$.

Among this $x$ numbers no number can have a factor that is greater than $x$.

Example: $30^2-30⋅29=900-870=30$. Out of these $30$ numbers no number can have a factor greater than $30$.

To count all the odd numbers that are prime and that are smaller or equal to any given $x$:

$y = pi (x) + 1$, where $pi (x)$ is the prime counting function and I deduct $-1$ because in this case I am not including $2$ as an odd prime.

In my example : $y = pi (30) – 1 = 10-1=9$.

To count all the odd numbers that are not prime:

$z = frac{1}{2}x – y$.

In my example : $z = frac {30}{2} -9 =15-9=6$.

The only problem is that we don’t know which of theses numbers are prime and which are not.

Assuming against reality and in an imaginary world, that all the prime numbers are the first numbers on the list that are greater than $1$, and all the remaining composite odd numbers are the last numbers on the list including the number $1$.

In our example:

Prime numbers: $3,5,7,9,11,13,15,17,19$.

Composite Numbers $1,21,23,25,27,29$.

Going back to the $x$ numbers that are between $x(x − 1)$ and $x^2$, in my example the numbers between $870$ and $900$

From here on I will carry on describing my question with my example:

Since I am pretending that all the first numbers greater than $1$ are prime, we need to use the inclusion exclusion principle (even on $9$ and $15$).

Let the answer be $A$ for how many of the $15$ odd numbers are not divisible by
$3,5,7,9,11,13,15,17,19$ (pretending they are all primes)?

$15 – (frac{15}{3} + frac{15}{5} + frac{15}{7} + frac{15}{11} + frac{15}{13} + frac{15}{17} + frac{15}{19}) + (frac{15}{15} + frac{15}{21} + frac{15}{27} + frac{15}{33} + frac{15}{39} + frac{15}{45} + frac{15}{51} + frac{15}{57} + frac{15}{35}+ frac{15}{45} + frac{15}{55} + frac{15}{65} + frac{15}{75} + frac{15}{85} + frac{15}{95} + frac{15}{63} + frac{15}{77} + frac{15}{91} + frac{15}{105} + frac{15}{119} + frac{15}{134} + frac{15}{99} + frac{15}{117} + frac{15}{135} + frac{15}{153} + frac{15}{171} + frac{15}{209} + frac{15}{143} + frac{15}{165} + frac{15}{187} + frac{15}{209} + frac{15}{195} + frac{15}{225} + frac{15}{255} + frac{15}{285} + frac{15}{323}) – …$

So far we have:

$A=15 – (14.332166285726656) + (8.075594188477142)-…$ and that is before I continue the inclusion exclusion principle.

I apologize for not continuing, I have tried for several hours and I just keep messing it up, and it wouldn’t matter for the remaining of my question:

As $x$ grows in this imaginary world, what will happen to $A$? for example: will it always stay positive?