I was recent reading through Paul Erdos's classic elementary proof of Sylvester-Schur. It occurred me that there is a simple argument that when $x$ is sufficiently large and if $p_i$ represents the $i$th prime such that $p_i \le n < p_{i+1}$, then there are least $n - i$ numbers in the sequence $x, x+1, \dots, x+n-1$ with prime divisors greater than $n$.

Have there been any well known result that goes beyond Sylvester-Schur in terms of the count of numbers with a prime divisor greater than $n$ in the integer sequence $x, x+1, \dots, x+n-1$? For example, trivially, $x > (n-1)!$ will have this property of at least $n - i$ numbers in the sequence $x,x+1,\dots,x+n-1$ with a prime divisor greater than $n$.

Edit: Here is the argument that I spoke of. Please let me know if anything is not clear.

Let $R(p,n)$ be the power of $p$ such that $p^{R(p,n)} \le n < p^{R(p,n)+1}$

Let $x > \prod\limits_{p < n}p^{R(p,n)}$ where $p$ is each prime less than $n$.

Let $p_i$ be the $i$th prime such that $p_i \le n < p_{i+1}$

Let $t_k$ be any integer $x \le x+t_k < x+n$ and $gpf(x+t_k) \le p_i$ where

*gpf*= greatest prime factor.

Claim 1: There exists a prime $q \le p_i$ such that $q^v \ge n$ and $q^v | x + t_k$

This follows directly from $x > \prod\limits_{p < n}p^{R(p,n)}$. If this were not true, then $x$ must necessarily be less than or equal to $\prod\limits_{p < n}p^{R(p,n)}$

Claim 2: There are at most $i$ such instances of $t_k$

Since there are only $i$ distinct primes less than or equal to $n$, it follows that if there are more than $i$ instances, then at least two must involve the same prime.

Assume that there exists $k_1 < k_2$ such that:

$0 < k_2 - k_1 < n$

$q^{v_1} | (x + k_1)$ and $q^{v_1} \ge n$ and $gpf(x+k_1) \le p_i$.

$q^{v_2} | (x + k_2)$ and $q^{v_2} \ge n$ and $gpf(x+k_2) \le p_i$

There exists integers $a_1 >0, a_2 > 0$ where:

$a_1(q^{v_1}) = x+k_1$ and $a_2(q^{v_2}) = x+k_2$

if $v_1 < v_2$, then:

$x+k_2 - x + k_1 = q^{v_1}[a_2(q^{v_2 - v_1}) - a_1]$

Now, $q^{v_1} \ge n$ and $[a_2(q^{v_2 - v_1}) - a 1] \ge 1$ which is impossible since $x + k_2 - x + k_1 < n$

if $v_1 > v_2$, then:

$x+k_2 - x + k_1 = q^{v_2}[a_2 - a_1(q^{v1-v2})]$ and the same argument applies.