I was recent reading through Paul Erdos's classic [elementary proof](https://www.renyi.hu/~p_erdos/1934-01.pdf) 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](http://mathworld.wolfram.com/GreatestPrimeFactor.html).


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.
> 
> 
> 



Going beyond the Sylvester and Schur theorem with regard to $x,x+1,\dots,x+n-1$
I like Larry Freeman's proof so much, I am going to rephrase it, first by weakening it.


Claim (Freeman): Given integer $n$ greater than 1, and given $x$ greater than $n!$, there are at most $\pi(n)$ many $n$-smooth integers in the interval $[x+1,x+n]$.


Indeed, his proof is an injection from the subset of $[x+1,x+n]$ which have largest prime factor at most $n$ into the set of primes in $[1,n]$.


Proof sketch: for each smooth number, pick the largest prime power factor of that number, and map the number to that prime. If two numbers map to the same prime, their difference is divisible by a prime power less than $n$. But then (as Larry points out), one of these smooth numbers must be less than $n!$, as it is then a divisor of $n!$. End of proof sketch.


The sketch above also works with Larry's version where $x$ is allowed to be as small as the least common multiple of the first $n$ numbers. I imagine this is folklore, and appears somewhere in the literature on smooth numbers. I would like to see a reference to this result, as well as improvements.


Fixing $n$, the $n$-smooth numbers get increasingly sparse, and the expected count of these numbers in a small interval decays, probably exponentially. Besides a result of Stormer on consecutive smooth numbers, I am not aware of a result like Larry's that is phrased in a combinatorial fashion. I hope others can show us such references. Even data showing the decay for small values of n would be welcome.


**Edit 2017.02.17 GRP**


'Prime factors of arithmetic progressions and binomial coefficients' is the title of a paper by Shorey and Tijdeman which has an argument similar to the above. This paper has some answers to the question as well as an extensive bibliography.


**End Edit 2017.02.17 GRP**


**EDIT 2018.01.27 GRP**


Many thanks to Jose Brox who in helping with figuring out how primes jump found the following paper:


MICHEL LANGEVIN
Plus grand facteur premier d’entiers en progression arithmétique


Séminaire Delange-Pisot-Poitou. Théorie des nombres, tome 18, no 1 (1976-1977),
exp. no 3, p. 1-7


in which a nice extension is proved. The interval of n numbers can be an arithmetic progression of n terms with common difference coprime to all terms, with the restriction that all terms are positive and none divide lcm[1..n], and the same map will be injective. The proof is almost the same as above.


Being annoyed at having to leave out divisors of lcm[1..n], I found an argument which allows one to include one divisor, at the possible cost of altering the map at that divisor. It is a lot of work for small gain, but it uses a weak combinatorial estimate on the number of certain perfect powers in an interval, so I am pleased with it. I can for large n bring the forbidden divisors down by a factor of square root of n, and hope to improve this further, using just combinatorial arguments.


**END EDIT 2018.02.27 GRP**


Gerhard "Too Long For A Comment" Paseman, 2017.02.16.


