# Counter example for a divisibility relation

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:

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

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

• I have tested it up to $n=2000$. – Adam Jun 10 '18 at 11:26

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.
• @Adam The added observations are trivial for similar reasons (e.g., the $\mathfrak d(\ldots)$ expression is trivially a divisor of $n$) – Hagen von Eitzen Jun 10 '18 at 12:34