Because I am yet to see a concise theoretical foundation as to why it may well be true for all natural number pairs $(n,k)$ greater than $1$, I have been seeking a counter example to disprove that:

$$\gcd\Bigl(\Bigl\lfloor\Bigl(\frac{p_n}{n}\Bigr)^{\frac{1}{k+2}}\Bigr\rfloor,\lfloor n^{\frac{1}{k+2}}\rfloor\Bigr)=1 \quad\quad\forall\,n,k \in \mathbb N \,\backslash\,{\{1}\}$$

Where $p_n$ is the $n^{th}$ prime number.

However this has proved to be quite infrequent and only possible at considerably large values of $n$.

Does anyone have any advice as to another approach in disproving the above lemma?

As far as proof is concerned, well, it only makes intuitive sense that it may be true in consideration of identities I have previously established which can be reviewed here

But as you will see, there is clearly some huge gaps if indeed a concise derivation exists.

thankyou for helping in advance.


1 Answer 1


Even if $k$ is only $2$, then $\bigg(\dfrac{p_n}{n}\bigg)^{\frac{1}{k+2}}$ will be less than $2$ until $\dfrac{p_n}{n} \ge 16$. That doesn't happen until $n$ is well over a million, so a large value of $n$ is indeed required.

A value of $n$ that will work is $6^8 = 1679616$; then $p_n = 26941241$ and $\dfrac{p_n}{n} \approx 16.040119$. Thus, with $k=2$, $$ \gcd\Bigl(\Bigl\lfloor\Bigl(\frac{p_n}{n}\Bigr)^{\frac{1}{k+2}}\Bigr\rfloor,\lfloor n^{\frac{1}{k+2}}\rfloor\Bigr)=\gcd(2,36)=2. $$

  • $\begingroup$ did your computation there involve something to do with having expressed your value as $6^8$ out of curiosity? $\endgroup$ Mar 29, 2019 at 3:41
  • 1
    $\begingroup$ For slightly smaller $n$, $\lfloor n^{\frac{1}{k+2}}\rfloor$ was $35$; I chose $6^8$ to push it up to an even value. $\endgroup$
    – FredH
    Mar 29, 2019 at 3:45

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