Minimum difference between consecutive multiples of $k$ that are $k\text{-rough}$

Question

A $k$-rough integer is any integer whose prime factors are all greater than or equal to $k$.

Is there a known formula for the smallest possible difference between consecutive $k$-rough multiples of $k$ when $k$ is prime? I've so far been unable to produce or locate one.

Answer 1

For $k=2$ the answer is $2$.

For prime $k\ge3$, $k$-rough multiples of $k$ must be odd so must differ by at least $2k$.

Let $k\#$ be the product of all primes up to and including $k$. Then $k\#-k$ and $k\#+k$ are both $k$-rough, multiples of $k$, and differ by $2k$.