# Prime divisor in positive integers sequences

I would like to know if anyone has an ideea if the following statement is true. For any sequence of consecutive positive integerers $$(n_0, n_0+1,..., n_0+k).$$ Where $$n_0 \ge 1, k\ge 0,$$ but $$k\ge 1$$ if $$n_0 = 0$$ (so different from the sequence (1)). There exists a prime number p that divides only one number in the sequence.

For example in the sequence (14, 15, 16) 7 divides only one of them, it doesn't have to be unique as 5 also divides only one of them.

A few restrictions that I managed to find for a counter-example to exist:

• The sequence cannot contain a prime number, otherwise it will also contain a largest prime number and that prime will only divide itself.
• k has to be smaller than $$2\times n_0,$$ otherwise the sequence will contain a prime number.
• The number of elements in the sequence (i.e. $$k+1$$) cannot be a prime number, otherwise that prime will only divide one number.

I made a simple computer program to check for such a sequence and couldn't find any in all possible sequences for up to $$n_0 = 200000.$$

• Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. Oct 11, 2018 at 7:36
• See here: en.wikipedia.org/wiki/… Oct 11, 2018 at 9:54
• Well doesn't Sylverster's theorem refers to a sequence of numbers that are less than 2(k+1)? It's just another version of the postulate and I did state that for such a sequence to exists k has to be smaller than 2 × n0 Oct 11, 2018 at 11:29
• It solves a part of the problem, but the numbers have to all be greater than k. It does make it extremely restrictive for such a sequence to exist, but I can't quite put my finger on why it can't exist. Oct 11, 2018 at 11:46

If $$n_0 > k+1$$ we have Sylvester's Theorem (see here, see here for the original paper by Sylvester and see here for a proof by Erdős), which states that $$n_0(n_0 + 1) \cdots (n_0 + k)$$ is divisible by a prime $$p$$ bigger than $$k+1$$, so that exactly one element of our sequence is divisible by $$p$$. For $$n_0 \le k+1$$ we have a prime $$p$$ with $$n_0 \le \left \lceil(n_0 + k)/2 \right\rceil < p \le n_0 + k$$ by Bertrand's Postulate and no other integer in the sequence can be divisible by $$p$$, since $$2p > n_0 + k$$.