$p$-simple integers from between $n$ and $n+p-1$

Question

Let $\ p\ $ be an arbitrary prime. Then an integer $\ s\ $ is called $p$-simple $\ \Leftarrow:\Rightarrow\ s\ $ is not divisible by any prime $\ q<p.\ $

Could you prove my conjecture (or is it known one way or another?):

For every prime $\ p\ $ and for every every integer $\ n\ $ there exists a $p$-simple integer $\ s\ $ such that $\ n\le s < n+p$.

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NOTES:

• There is a $p$-simple integer such that $\ n\le s<n+p\ $ for every $\ n\ $ such that $\ -2\cdot p < n \le p$.

• If the conjecture holds restricted to all positive integers $\ n\ $ then it holds in full, for all integers $\ n$.

• There is a $p$-simple integer $\ s\ $ such that $\ p+1 \le s\le 2\cdot p\ $ (Chebyshev--actually, $\ s\ $ can be a prime).

• The conjecture holds for $\ n=2\ $ (trivial); and for $\ n=3\ $ (at least one of the three consecutive integers is odd); and for $\ p=5\ $ (among any $5$ consecutive integers, at the most $3$ are even, at the most $2$ are divisible by $3$, whole one of the even ones is divisible by $3$ when there are there three even numbers--thus one of these consecutive $5$ integers is not divisible by $2$ nor by $3$).

Answer 1

The conjecture is false. Rankin (1938) proved that there exists a constant $c>0$ such that for all $x>20$, there exist at least $$ c x\frac{(\log x)(\log\log\log x)}{(\log\log x)^2} $$ consecutive integers, each of which are divisible by some prime less than $x$. Note that the fraction here tends to infinity as $x\to\infty$. More recently, Ford-Green-Konyagin-Maynard-Tao (2014) improved the denominator $(\log\log x)^2$ above to $\log\log x$, and this is the state-of-the-art.

On the positive side, Sylvester (1892) proved that for any $n>p$, there is an integer $n\leq s<n+p$, which has at least one prime factor exceeding $p$. A simple proof was given by Erdős (1934).

Answer 2

Given a prime $p$, let $P$ be the product of the primes less than $p$. A $p$-simple integer $s$ is then an integer satisfying $\gcd(P,s)=1$ (so $s$ is a totative of $P$), and the posted conjecture asserts that any interval of $p$ consecutive integers contains at least one such $s$, and this means the difference between two consecutive totatives of $P$ is at most $p$.

This holds for $p\leq 11$ (as $g(210)=10$), but not for larger $p$, as the example in the comments with $p=13$ and $n=114$ shows. The Jacobsthal function $g(m)$ which measures the maximal difference between consecutive totatives of $m$ is larger than $k=\Omega(m)$, the number of distinct prime factors of $m$, and when restricted to primorials $P$ is easily shown to grow at least as fast as $2q$, where $q$ is the largest but one prime less than $p$. As mentioned in another answer, Rankin (and earlier, Westzynthius in 1931) showed that for any positive $C$, there are some $m$ with $g(m) \gt Ck\log k$.

Gerhard "Now I've Mentioned Jacobsthal's Function" Paseman, 2016.06.16.