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?):



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> 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 $3$).



$p$-simple integers from between $n$ and $n+p-1$
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)](http://arxiv.org/abs/1412.5029) 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)](http://www.renyi.hu/~p_erdos/1934-01.pdf).


