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Is it true that the sum of a specific floor functions function of a prime = 1?

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:

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

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

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
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Is it true that $\sum_{i | p\\#}^{p\\#} \floor{\frac{p}{i}\mu(i)\rfloor the sum of a floor functions of a prime = 1$1?
show/hide this revision's text 1

Is it true that $\sum_{i | p\\#}^{p\\#} \floor{\frac{p}{i}\mu(i)\rfloor = 1$

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:

$\frac{2}{1} - \frac{2}{2} = 1$

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

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