This one seems counter intuitive to me but I am not seeing a mistake in my reasoning.
Please let me know if you find one.
Let:
- $x > 0$ be an integer
- $\mu(x)$ be the möbius function
- $x\#$ be the primorial for $x$
- $r(m,d)$ be the remainder when $d$ divides $m$.
Does this follow:
$1 + \sum\limits_{i|x\#}\dfrac{r(x,i)}{i}\mu(i) = \prod\limits_{p\text{ prime, } p \le x}\left(\dfrac{p-1}{p}\right)x$
Here's my thinking:
(1) $\sum\limits_{i|x\#}\left\lfloor\dfrac{x}{i}\right\rfloor\mu(i) = 1$
This follows from the inclusion-exclusion principle and logic found in step 1 here
(2) $1 + \sum\limits_{i|x\#}\dfrac{r(x,i)}{i}\mu(i) = \sum\limits_{i|x\#}\left(\dfrac{x}{i}\right)\mu(i)$
$1 + \sum\limits_{i|x\#}\dfrac{r(x,i)}{i}\mu(i) = \sum\limits_{i|x\#}\left(\left\lfloor\dfrac{x}{i}\right\rfloor+\dfrac{r(x,i)}{i}\right)\mu(i)$
$= \sum\limits_{i|x\#}\left(\dfrac{x}{i}-\dfrac{r(x,i)}{i} +\dfrac{r(x,i)}{i}\right)\mu(i)$
$=\sum\limits_{i|x\#}\left(\dfrac{x}{i}\right)\mu(i)$
(3) $\sum\limits_{i|x\#}\left(\dfrac{x}{i}\right)\mu(i) = \prod\limits_{p\text{ prime, } p \le x}\left(\dfrac{p-1}{p}\right)x$
$\sum\limits_{i|x\#}\left(\dfrac{x}{i}\right)\mu(i) =\prod\limits_{p\text{ prime, } p \le x}\left(1 - \dfrac{1}{p}\right)x$
$= \prod\limits_{p\text{ prime, } p \le x}\left(\dfrac{p}{p} - \dfrac{1}{p}\right)x = \prod\limits_{p\text{ prime, } p \le x}\left(\dfrac{p-1}{p}\right)x$
As has been stated in a comment, your calculations are correct.