Does the Euler product stand for $a(n)=rad(n)$? Or more generally, for multiplicative functions which are not completely multiplicative?
Where rad is the product of a number's distinct prime factors.
Wikipedia states that we can calculate the Euler product where $a(n)$ is multiplicative but doesn't stipulate completely multiplicative.
Rad function is multiplicative but not completely multiplicative.
So we would have:
$\displaystyle\sum_{n=0}^{\infty}rad(n)n^{-s}=\prod_p \left(1+rad(p)p^{-s}+rad(p^2)p^{-2s}+rad(p^3)p^{-3s}+\ldots\right)\\=\prod_p 1+p^{1-s}+p^{1-2s}\ldots=\prod_p \left(p(1+p^{1-s}+p^{1-2s}\ldots)+1-p\right)$
... there's a bit more work needed to relate this back to $\zeta(s)$ but I want to be sure the Euler product is valid first.