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 $n_0 = 200000.$