Let:
- $x>0, n >0$ be integers
- gcd$(s,t)$ be the greatest common divisor of $s$ and $t$
- $\mu(x)$ be the möbius function
For $x \le n$ and gcd$(x,30)=1$, the count is:
$$\sum_{i | 30}\left\lfloor\frac{n}{i}\right\rfloor\mu(i) = n - \left\lfloor\frac{n}{2}\right\rfloor - \left\lfloor\frac{n}{3}\right\rfloor - \left\lfloor\frac{n}{5}\right\rfloor + \left\lfloor\frac{n}{6}\right\rfloor + \left\lfloor\frac{n}{10}\right\rfloor +\left\lfloor\frac{n}{15}\right\rfloor - \left\lfloor\frac{n}{30}\right\rfloor$$
For $x \le 30$ and gcd$\left(x(x+2\right),30)=1$, the count is:
$$30 - \left\lfloor\frac{30}{2}\right\rfloor - 2\left\lfloor\frac{30}{3}\right\rfloor - 2\left\lfloor\frac{30}{5}\right\rfloor + 2\left\lfloor\frac{30}{6}\right\rfloor + 2\left\lfloor\frac{30}{10}\right\rfloor +2\left\lfloor\frac{30}{15}\right\rfloor=3$$
I am not clear how to represent this using the möbius function.
I had expected this to work:
$$\sum_{i|30}\left\lfloor\frac{n}{i}\right\rfloor\mu(i) + \sum_{i|30\text{ and }i>2}\left\lfloor\frac{n}{i}\right\rfloor\mu(i)$$
It does not work for $n=30$. I am not clear why, in this case, $-2\left\lfloor\frac{n}{30}\right\rfloor$ is not needed.
How do you correctly count the integers $x \le n$ where gcd$\left(x(x+1),30\right)=1$ using the möbius function?
Edit:
rtybase makes a good point. This can be precisely calculated using:
$$\left\lfloor\frac{n+19}{30}\right\rfloor + \left\lfloor\frac{n+13}{30}\right\rfloor + \left\lfloor\frac{n+1}{30}\right\rfloor$$