I apologize if this is a naive question about greatest prime factors (gpf). I was thinking about the sequence of integers where $gpf(x) \le p$ where $p$ is any prime.

Clearly, as $x$ increases, the distance $d$ between an integer where $gpf(x) \le p$ and $gpf(x+d) \le p$ increases at a seemingly every increasing rate.

For all primes $p$, does there exist an integer $C$ where if $x ≥ C$, then there is at most $1$ integer in the sequence $x+1, x+2, \dots, x+p$ has $gpf(x) \le p$

For example, if $p=2$, $C = 2$ since for any $x \ge 2$, either $x$ or $x+1, gpf(x) > 2$

I apologize if this is a naive question about greatest prime factors (gpf). I was thinking about the sequence of integers where $\mathrm{gpf}(x) \le p$ where $p$ is any prime.

Clearly, as $x$ increases, the distance $d$ between an integer where $\mathrm{gpf}(x) \le p$ and $\mathrm{gpf}(x+d) \le p$ increases at a seemingly every increasing rate.

For all primes $p$, does there exist an integer $C$ where if $x \ge C$, then there is at most one integer in the sequence $x+1, x+2, \dots, x+p$ has $\mathrm{gpf}(x) \le p$

For example, if $p=2$, $C = 2$ since for any $x \ge 2$, either $x$ or $x+1, \mathrm{gpf}(x) > 2$