Is there a counterexample ( other than $n=1$) for this divisibility relation:

$$\vartheta _{{n}}=\min(\mathcal D(n\cdot\bigl\lfloor \frac{p_n}{n} \bigr\rfloor) \backslash {\{1}\})$$

$${\Biggl\{\frac {\bigl\lfloor \sqrt {n} \bigr\rfloor\gcd(n ,\vartheta _{{n}})}{\gcd(\bigl\lfloor \sqrt {n} \bigr\rfloor ,\vartheta _{{n}})}\Biggr\}}=0\quad\quad\quad\quad\quad\quad\quad\quad\quad(A0)$$


$\mathcal D(n)$ is the set of all divisors of $n$

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

${\{x}\}$ is the fractional part of $x$

The following are additional problems that require explicit proof in order for the first to be considered true:

The substitution of $n$ and $\bigl\lfloor \sqrt {n} \bigr\rfloor$ aside from the minimum divisor term holds:

${\Biggl\{\frac {n \cdot \gcd(\bigl\lfloor \sqrt {n} \bigr\rfloor\ ,\vartheta _{{n}})}{\gcd(n ,\vartheta _{{n}})}\Biggr\}}=0$ $ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad (A1)$

The factorials as follows are divisible:

${\Biggl\{\frac { \left( n-2 \right) !}{ \left( \bigl\lfloor \sqrt {n} \bigr\rfloor\ \right) !}}\Biggr\}=0 \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(A2) $

And this was what seems to be why that is so for the general case, but I may terribly embarass myself here again because I don't have the explicit proof:

${\Biggl\{\frac { k !}{ \left( \bigl\lfloor k^{n/m}\bigr\rfloor\ \right) !}}\Biggr\}=0 \land k \geq 1 \Rightarrow n \leq m\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad (A3.1)$

${\Biggl\{\frac { \left( \bigl\lfloor k^{n/m}\bigr\rfloor\ \right) !}{ k !}}\Biggr\}=0 \land k \geq 1 \Rightarrow n \geq m\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad (A3.2)$

  • $\begingroup$ I have tested it up to $n=2000$. $\endgroup$ Jun 10, 2018 at 11:26

1 Answer 1


For $n>2$, we already have $$ 2\gcd(\lfloor\sqrt n\rfloor, \ldots)\mid \phi(n)\lfloor \sqrt n\rfloor$$ because the gcd of two numbres divides each of these two numbers and because $\phi(n)$ is even.

  • $\begingroup$ Correct yes I should have simplified this further I'm sorry for that, I have too many problems that are very similar to one another, some of which involve only the numerator and denominator of the Euler product, and this follows from those and I hadn't thought of this yet $\endgroup$ Jun 10, 2018 at 11:43
  • $\begingroup$ I added the relevant information so you can see why I had the totient present in the first sir $\endgroup$ Jun 10, 2018 at 11:56
  • $\begingroup$ @Adam The added observations are trivial for similar reasons (e.g., the $\mathfrak d(\ldots)$ expression is trivially a divisor of $n$) $\endgroup$ Jun 10, 2018 at 12:34
  • $\begingroup$ True, the two gcd terms not being divisible can only mean that is true for a n $\endgroup$ Jun 10, 2018 at 12:46
  • $\begingroup$ When you say trivially, what is the difference between a non trivial and trivial divisor? $\endgroup$ Jun 10, 2018 at 12:47

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