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 $n_0 = 200000.$


1 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 $2p > n_0 + k$.

  • $\begingroup$ Great answer! Can you please add the link to Sylvester's Theorem to your answer so that everything's in the same place? $\endgroup$ Oct 11, 2018 at 13:33
  • $\begingroup$ Thank you very much for the answer ! Would you happen to have a demonstration for Sylvester's Theorem. He has few of them but didn't manage to find anything about it. $\endgroup$ Oct 11, 2018 at 20:09
  • $\begingroup$ I have added a link to Sylvester's paper from 1892. It is not easy by any means, but I find his writing style beautiful! $\endgroup$
    – Woett
    Oct 11, 2018 at 21:55

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