# 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 $$n_0 = 200000.$$

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 $$2p > n_0 + k$$.