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,

-Larry

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

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

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,

-Larry