I noticed that for primes $p \le 109$, the following seems to be true:


$$\sum\_{i | p\#}^{p\#} \left\lfloor{\frac{p}{i}\mu(i)}\right\rfloor = 1$$


where $\mu(i)$ is the Mobius function.


For example: 


$\frac{2}{1} + \frac{2}{2}(-1) = 1$


$\frac{3}{1} + \lfloor\frac{3}{2}(-1)\rfloor + \frac{3}{3}(-1) + \lfloor\frac{3}{6}\rfloor = 1$


and so on...


I verified this up to $p=109$ using a simple java application.


I might be making a mistake in my code or my thinking. This seems like a very surprising result to me.


Is it correct? If it is, does it stop being true for some prime? Could anyone help me to understand this result.


Thanks very much,


-Larry



Is it true that the sum of a specific floor function of a prime = 1?
Assuming that $p\text{#}$ means the product over primes $\prod\_{q\leq p}q$, then it is clear that 
$$\sum\_{i\leq p} \mu(i)[\frac{p}{i}] = \sum\_{i | p\text{#}} \mu(i)[\frac{p}{i}].$$


But this formula is very well known:
$$\sum\_{d\leq n} \mu(d)[\frac{n}{d}]=\sum\_{k\leq n}\sum\_{d | k} \mu(d)=1+0+0+\ldots$$


