Is there a counterexample ( other than $n=1$) for this divisibility relation:
$$\vartheta _{{n}}=\min(\mathcal D(n\cdot\bigl\lfloor \frac{p_n}{n} \bigr\rfloor) \backslash {\{1}\})$$
$${\Biggl\{\frac {\bigl\lfloor \sqrt {n} \bigr\rfloor\gcd(n ,\vartheta _{{n}})}{\gcd(\bigl\lfloor \sqrt {n} \bigr\rfloor ,\vartheta _{{n}})}\Biggr\}}=0\quad\quad\quad\quad\quad\quad\quad\quad\quad(A0)$$
Where:
$\mathcal D(n)$ 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:
${\Biggl\{\frac { k !}{ \left( \bigl\lfloor k^{n/m}\bigr\rfloor\ \right) !}}\Biggr\}=0 \land k \geq 1 \Rightarrow n \leq m\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad (A3.1)$
${\Biggl\{\frac { \left( \bigl\lfloor k^{n/m}\bigr\rfloor\ \right) !}{ k !}}\Biggr\}=0 \land k \geq 1 \Rightarrow n \geq m\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad (A3.2)$
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.