I am working on a problem in commutative ring theory, that deals with $p$-adic valuations. This leads to a number theoretical question that I want to explain in the following.


Let $n \in \mathbb{N}$ and $k$ an integer $\leq n/2$. Then, by the well-known result of Sylvester, there is an integer in $\{n, n-1,..., n-k+1\}$ that has a prime factor $>k$.


There is a very practical survey by Shorey and Tijdeman on generalizations of this theorem. For instance, it is known that the maximal distance of two positive integers having a prime factor $>k$ is $\leq \Big(\frac{1}{2} + o(1)\Big)\frac{k}{\log(k)}$. In particular, for large enough $k$ this difference is less than $\pi(k) < k$.


There are some similar results, but none of them helped me with my actual question, although they point in the direction of a positive answer to it:


**Question**: Can one explicitely give a constant $C$ with the following property?: For each positive integer $n \geq C$ that is not a prime power and for each integer $k$ with $1<k\leq n/2$, there are two *different* $a,b \in \{n,n-1,...,n-k+1\}$ having a prime factor $>k$.


Thank you in advance for any help!



Strengthening Sylvester's theorem
With motivation from the original poster, a key idea from Sylvester, and technical inspiration from Iosif Pinelis, I contribute an observation that helps toward an answer.


I use m instead of n and n instead of k. I start with the inequality that p! is strictly less than 3^p for p less than 7, and less than (p/2)^p for all larger integers p. We will set p =$\pi$(n).


Consider the product of integers in (m,m+n], and write it as W(n!)B, where W are all the prime factors (with multiplicity) at most n dividing (m+n)!/(m!n!), leaving B as the product of the remaining prime factors which are all larger than n, and B=1 if there are no such large prime factors.


Sylvester's observation is that W is at most (m+n-p+1)...(m+n). If B=1 the interval (m,m+n] has all numbers being n-smooth. The extended observation (which I think is new and hopefully orignal) is that WB is at most (m+n-p-d+1)...(m+n) if there are at most d many numbers in (m,m+n] which are not n-smooth. We fix d and observe that the original problem relates to d=1 in what follows.


Under supposition that there are not d+1 many non smooth numbers in (m,m+n], we now have (n!) Is at least (m+1)...(m+n-p-d). Write m as kn + i for positive integer k and non negative integer i (choose i less than n for less confusion). We now have (p+d)! Is at least (and for large enough n strictly greater than) k^(n-p-d).


So if (m,m+n] has at most d numbers which are not n-smooth, then we use the inequality above to note that when p+d is greater than 6, k is strictly less than ((p+d)/2)^((p+d)/(n-(p+d))). To save on plus signs, write q=p+d.


By the above, when q is at most 6 and n is at least 2q, then k is at most 2. (I leave the case n smaller than 12 and arbitrary d to the reader.) As n grows, q(1+ log (q/2)) will be less than n (because d is fixed), and one can use current literature or supercomputers to compute for which n this holds, in which case k is strictly less than e.


So given d, one can compute n0 without much challenge to find that (m,m+n] has d+1 non smooth numbers for n greater than n0 and for m at least as large as 3n.


To handle the remaining case for small d (d less than 6), use Nagura or similar as outlined in the other answer of mine to find d+1 non smooth integers in the interval for when m is in [n,3n). This should hold for m at least 150, giving C is less than 150.


Gerhard "Would James Joseph Be Approving?" Paseman, 2020.06.01.


