Prime divisor in positive integers sequences

Question

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 $n_0 = 200000.$

Answer 1

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 $2p > n_0 + k$.