Take the 2-minute tour ×  I noticed that for primes $p \le 109$,
the following seems to be true:  $\sum_{i | p\\#}^{p\\#}
\lfloor{\frac{p}{i}\mu(i)}\rfloor = 1$  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,   share|improve
this question In your example, you seem to be summing
mu(i)*floor(p/i) rather than floor(mu*p/i). –  David Cohen Jun
20 '12 at 22:59 Thanks, David. I'll update the example. –  Larry
Freeman Jun 20 '12 at 23:01  1 Answer 1  up vote 6 down vote
accepted  Assuming that $p\text{#}$ means the product over
primes $\prod_{q\leq p}q$, then it is clea