Counting consecutive integers that are divisible by primes relatively prime to an arbitrary $n$ Let:

*

*$c > 0, n, m, x > 0$ be an integers

*$p\#$ be the primorial of $p$

*$D_n(m,x)$ be the count of integers $i$ where:



*

*$m-x \le i < m$

*There exists a prime $p$ that $p \nmid n$ but $p | i$

It occurs to me that for any $c$, there exists $n,m,x$ such that:
$$D_n(m,x) \ge D_n(0,x) + c$$
Here is the argument:
(1)  From well-known properties [see (20)] of Bertrand's Postulate, there exists $x$ for any such $c$ such that:
$$\frac{x}{2} < p_0 < p_1 < p_2 < \dots < p_c < x$$
where each $p_i$ is a prime
(2)  $D_{p_0\#}(0,x) = c$
(3)  Using the Chinese Remainder Theorem, there exists $u$ such that:

*

*$p_c | u$

*$p_{c-1} | u+1$

*$p_{c-2} | u+2$

*$\vdots$

*$p_1 | u+c-1$
(4)  Let $m = x + u$
(5)  It follows that:
$$D_{p_0\#}(m,x) = 2c$$
Is my reasoning correct?  Did I make a mistake?

Edit:  Fixed a minor mistake identified by John Omielan
 A: What you've done is correct. I have just a couple of suggestions, and you made a minor mistake.
For your ($1$), an alternative is to use that the prime number theorem shows the prime gaps eventually have an upper bound which is an arbitrarily small multiplier of the prime number. As stated in the Upper bounds section of Wikipedia's "Prime gap" article,

For every $\epsilon \gt 0$, there is a number $N$ such that for all $n \gt N$

$\; \; \; \; g_{n} \lt p_{n}\epsilon$.

You can just choose $\epsilon = \frac{1}{c + 1}$ in your case.
The other suggestion is you could have explicitly mentioned that, for each $p_i$, there is also another integer $u + (c - i) + p_i$ which is less than $x$, is divisible by $p_i$, and is not divisible by any $p_j$ where $j \neq i$ (so it won't be double-counted). This means it's an additional integer to include in the $D_n(m,x)$ count, which then explains how you got the extra $c$ to get $D_{p_0\#}(m,x) = 2c$.
As for the small mistake, your $p_1 | u + c$ should be $p_1 | u + c - 1$ instead.
When I determined the counter-example for your related question Reasoning about relatively prime factors of consecutive integers, I thought it'll also be the case for any integer larger than $1$, i.e., $c$ in your question here. I didn't consider then how to possibly prove it, but I'm glad to see you've done it here with your post.
