Sizes of Blocks of Consecutive Integers Divisible by at Least One Prime Less than or Equal to $r$.

Question

Let $f(r)$ be the largest integer such that there exists a block of $f(r)$ consecutive integers each divisible by some prime that is less than or equal to $r$. For example, $f(2)=1$ because it is impossible to find two consecutive integers each divisible by $2$. Similarly, $f(3)=3$ because it is impossible to find four consecutive integers all divisible by $2$ or $3$, but $2,3,4$ is a block of $3$ such consecutive integers. I am curious about the existence of an upper bound on $f$. I suppose some basic sieve method could come into play, and I am sure I could get a weak bound with rough approximations. However, I was wondering if there is any good upper bound already in existence. Thank you.

Answer 1

What you are looking for is the sequence A058989, which is also $j(n\#)-1$, one less than the Jacobsthal function of the primordial numbers. There are some known asymptotic upper and lower bounds for this function: $$ (2e^\gamma + o(1)) \frac{n \log^2 n \log \log \log n}{(\log \log n)^2} < j(n\#) \ll n^2 \log^2 n $$

this is related to some interesting properties of prime gaps, for example from the left had side you can derive that there are prime gaps substantially larger than the average.