Does the following sum have a closed-form expression? I've tried an Inclusion-Exclusion interpretation, to no avail:

$f(n, p) = \sum_{i = 1} ^ n \lfloor \frac{n}{i} \rfloor \phi(p i) (-1) ^ i$

($p$ is a prime number.)

Interesting observation: for $p > n$, $f(n, p) / (p - 1)$ appears to be independent of $p$.

Does the following sum have a closed-form expression? I've tried an Inclusion-Exclusion interpretation, to no avail:

$f(n, p) = \sum_{i = 1} ^ n \lfloor \frac{n}{i} \rfloor \phi(p i) (-1) ^ i$

($p$ is a prime number.)

Interesting observation: for $p > n$, $f(n, p) / (p - 1)$ is independent of $p$.

Does the following sum have a closed-form expression? I've tried an Inclusion-Exclusion interpretation, to no avail:

$\sum_{i = 1} ^ n \lfloor \frac{n}{i} \rfloor \phi(p i) (-1) ^ i$

($p$ is a prime number.)

Does the following sum have a closed-form expression? I've tried an Inclusion-Exclusion interpretation, to no avail:

$f(n, p) = \sum_{i = 1} ^ n \lfloor \frac{n}{i} \rfloor \phi(p i) (-1) ^ i$

($p$ is a prime number.)

Interesting observation: for $p > n$, $f(n, p) / (p - 1)$ appears to be independent of $p$.