# Are there always at least $3$ integers $x$ where $an < x \le an+n$ and $\gcd(x,\frac{n}{4}\#)=1$

The answer seems to be yes.

Please let me know if I have made a mistake in my argument or if there is an opportunity to make the argument simpler, clearer, or more standard.

Let $a \ge 1, n \ge 2$ be integers.

Let $\frac{n}{4}\#$ be the primorial for $\left\lfloor\frac{n}{4}\right\rfloor$

Let $\gcd(a,b)$ be the greatest common divisor for $a$ and $b$.

Let $f(m,p)$ be the number of integers $x$ where $x \le m$ and $\gcd(x,p\#)=1$ so that $f(10,2) = 5$

Let $g(h,l,p)$ be the number of integers $x$ where $l < x \le h$ and $\gcd(x,p\#)=1$ so that:

$$g(an+n,an,\frac{n}{4}\#) = f(an+n,\frac{n}{4}\#) – f(an,\frac{n}{4}\#)$$

From the inclusion-exclusion principle, it follows (see explanation found here):

$$\left\lfloor\left(\prod\limits_{p_i \le \frac{n}{4}}\frac{p_i-1}{p_i}\right)m\right\rfloor \le f(m,\frac{n}{4}\#) \le \left\lceil\left(\prod\limits_{p_i \le \frac{n}{4}}\frac{p_i-1}{p_i}\right)m\right\rceil$$

So that:

$$g(an+n,an,\frac{n}{4}\#) \ge \left\lfloor\left(\prod\limits_{p_i \le \frac{n}{4}}\frac{p_i-1}{p_i}\right)(an+n)\right\rfloor – \left\lceil\left(\prod\limits_{p_i \le \frac{n}{4}}\frac{p_i-1}{p_i}\right)(an)\right\rceil \ge$$

$$\left\lfloor\left(\prod\limits_{p_i \le \frac{n}{4}}\frac{p_i-1}{p_i}\right)(an+n)\right\rfloor – \left\lfloor\left(\prod\limits_{p_i \le \frac{n}{4}}\frac{p_i-1}{p_i}\right)(an)\right\rfloor – 1 \ge$$

$$\left\lfloor\left(\prod\limits_{p_i \le \frac{n}{4}}\frac{p_i-1}{p_i}\right)(an+n) – \left(\prod\limits_{p_i \le \frac{n}{4}}\frac{p_i-1}{p_i}\right)(an)-1\right\rfloor = \left\lfloor\left(\prod\limits_{p_i \le \frac{n}{4}}\frac{p_i-1}{p_i}\right)(an+n – an)-1\right\rfloor =$$

$$\left\lfloor\left(\prod\limits_{p_i \le \frac{n}{4}}\frac{p_i-1}{p_i}\right)(n)-1\right\rfloor > \left\lfloor\left(\prod\limits_{2 \le w \le \frac{n}{4}}\frac{w-1}{w}\right)(n)-1\right\rfloor$$

Now, it seems to me that:

$$\prod\limits_{2 \le w \le \frac{n}{4}}\frac{w-1}{w}= \left(\frac{1}{2}\right)\left(\frac{2}{3}\right)\left(\frac{3}{4}\right)\dots\left(\frac{\left\lfloor\frac{n}{4}\right\rfloor-2}{\left\lfloor\frac{n}{4}\right\rfloor-1}\right)\left(\frac{\left\lfloor\frac{n}{4}\right\rfloor-1}{\left\lfloor\frac{n}{4}\right\rfloor}\right) = \frac{1}{\left\lfloor\frac{n}{4}\right\rfloor}$$

So that:

$$\left\lfloor\left(\prod\limits_{2 \le w \le \frac{n}{4}}\frac{w-1}{w}\right)(n)-1\right\rfloor = \left\lfloor\left(\frac{1}{\left\lfloor\frac{n}{4}\right\rfloor}\right)(n)-1\right\rfloor \ge 3$$

Edit 1: This argument is not valid.

Counter examples to this argument can be found here.

The conclusion may still be valid. I have an attempted a different argument to the same conclusion here.