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Can you provide a proof or a counterexample for the claim given below?

Inspired by Grimm's conjecture I have formulated the following claim:

Let $n_1,n_2,\dots,n_k$ be a sequence of $k$ consecutive odd numbers which are all composite. Let $\operatorname{gpf}(n_i)$ be the greatest prime factor of $n_i$. Then, all $\operatorname{gpf}(n_i)$, $1 \le i \le k$ are mutually different.

Try it yourself.

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  • $\begingroup$ "be a greatest" --> "be the greatest". also, what's $k$? $\endgroup$ Commented Oct 30, 2018 at 10:40
  • $\begingroup$ $3,5,7,9$. $3$ is the greatest prime factor of $3$ and $9$ $\endgroup$ Commented Oct 30, 2018 at 10:43
  • $\begingroup$ @mathworker21 $3,5,7$ are not composite numbers... $\endgroup$
    – Pedja
    Commented Oct 30, 2018 at 10:53
  • $\begingroup$ ah, my apologies. what is $k$ though? $\endgroup$ Commented Oct 30, 2018 at 10:54
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    $\begingroup$ If your conjecture is true, then for any prime $p$, there exists a prime between $p(p-2)$ and $p^2$. (There gpfs are $p$) Showing it is likely to be as hard as proving that there exists a prime between $n$ and $n+k\sqrt{n}$ for some constant $k$, which is open problem as of now. (Currently it's proven only for $n+n^{0.525}$) $\endgroup$
    – didgogns
    Commented Nov 1, 2018 at 15:38

1 Answer 1

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A conterexample: gpf(20449) = gpf(20475) = 13, and there are no prime numbers between them. According to my program's results, it is the only counterexample up to 1000000.

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