How can I evaluate ( estimate ) the sum $s(n)=\sum_{k=1}^{n-1}\mu ^{2}(k(n-k))$ ($\mu$ is Mobius function)  . Trivial estimate
$s(n)<\varphi (n)$ follow from the fact that $\mu (k(n-k))$ is zero if $k$ is not prime to $n$. It seem that in the case of $n$ - primorial, $s(n)$ is pretty close to $\varphi (n)$

How can I evaluate ( estimate ) the sum $s(n)=\sum_{k=1}^{n-1}\mu ^{2}(k(n-k))$ ($\mu$ is the Möbius function). Trivial estimate
$s(n)<\varphi (n)$ follows from the fact that $\mu (k(n-k))$ is zero if $k$ is not prime to $n$. It seem that in the case of $n$ - primorial, $s(n)$ is pretty close to $\varphi (n)$

How can I evaluate ( estimate ) the sum $s(n)=\sum_{k=1}^{n-1}\mu ^{2}(k(n-k))$ ($\mu$ is Mobius function) . Trivial estimate
$s(n)<\varphi (n)$ follow from the fact that $\mu (k(n-k))$ is zero if $k$ is not prime to $n$.

How can I evaluate ( estimate ) the sum $s(n)=\sum_{k=1}^{n-1}\mu ^{2}(k(n-k))$ ($\mu$ is Mobius function) . Trivial estimate
$s(n)<\varphi (n)$ follow from the fact that $\mu (k(n-k))$ is zero if $k$ is not prime to $n$. It seem that in the case of $n$ - primorial, $s(n)$ is pretty close to $\varphi (n)$