What is the Euler product for double summations?

Question

I know that the Euler product of a summation of multiplicative function is given by

$$\sum_nf(n)=\prod_p (1+f(p)+f(p^2)+....),$$

and if we have the Möbius function then it will be

$$\sum_n\mu (n)f(n)=\prod_p (1+f(p)).$$

I would like to know if there is a general formula for Euler product of a double summation? I saw in one of the papers he applied the Euler product like this

$$ \sum_{a,r}\mu (a)f(a)^2\mu (r)\frac{f(r)}{\sqrt{r}}=\prod_p (1+f(p)^2+\frac{f(p)}{\sqrt{p}})$$

I know that the Möbius function vanish all other terms, but I would like to know the general one without the Mobius. Thanks.

Answer 1

Sums of the form $S_0=\sum_{n,m}f(n)g(m)$ where $f,g$ have some multiplicative property should be doable by a double Euler formula. However it is more useful to have a multidimensional Euler formula for sums of the form $$S:=\sum_{(n_1,\ldots,n_k) \in \mathbb{N}^k}h(n_1,\ldots,n_k).$$ This can be done if the following two conditions hold. The first condition is that for all $$(n_1,\ldots,n_k),(m_1,\ldots,m_k) \in \mathbb{N}^k$$ with $$\gcd\Big(\prod_{i=1}^kn_i,\prod_{i=1}^km_i\Big)=1$$ one must have $$h(n_1m_1,\ldots,n_km_k)=h(n_1,\ldots,n_k)h(m_1,\ldots,m_k).$$ The second condition is that the infinite series $S$ must be absolutely convergent. If both of these conditions hold then Euler's product formula becomes $$S=\prod_p\Big(\sum_{(\nu_1,\ldots,\nu_k) \in \mathbb{Z}_{\geq 0}^k}h(p^\nu_1,\ldots,p^\nu_k)\Big).$$ This of course includes expressions like $S_0$ because one can take $k=2$ and $$h(n_1,n_2)=f(n_1)g(n_2).$$