TITLE: Is it possible to multiply two series to get as a result all composite numbers?
QUESTION [6 upvotes]: I was toying with the following problem:
Is it possible to find two infinite integer sequences $(a_n), (b_n)>0$ such that $\sum_{n=1}^{\infty}\frac{1}{(a_n)^s}\cdot \sum_{n=1}^{\infty}\frac{1}{(b_n)^s}=\sum_{n=1}^{\infty}\frac{1}{(c_n)^s}$ for every $s>1$? Here $c_n$ denotes the $n$-th composite number.
I can show that without loss of generality, $1\in a_n$ and for every $x$ with $\Omega(x)=2, x\in (b_n)$ but this did not help much. 
Can someone provide an answer to this problem?

REPLY [6 votes]: This is the answer of Greg Martin, with the correction of Mark Sapir, and details added.
Write $\Omega(n)$ for the number of prime factors of $n$ (counted with multiplicity), and $\Omega_{\operatorname{odd}}(n)$ for the number of odd prime factors of $n$ (counted with multiplicity), so $\Omega(n) = \Omega_{\operatorname{odd}}(n) + v_2(n)$ (where $v_2$ is the valuation at $2$).

Example. Let $A = \{1,2,4,8,\ldots\} = 2^{\mathbf Z_{\geq 0}}$, and let
  $$B = \{n\ |\ \Omega(n) = 2\} \cup \{n \text{ odd}\ |\ \Omega(n) \geq 3\}.$$
  For $n \in \mathbf Z_{>0}$, the number of representations $n = a \cdot b$ with $a \in A$ and $b \in B$ is $1$ if $n$ is composite, and $0$ otherwise.

Proof. Given $n \in \mathbf{Z}_{>0}$ composite (i.e. $\Omega(n) \geq 2$), define $k \in \mathbf Z_{\geq 0}$ as follows:

If $\Omega_{\operatorname{odd}}(n) \geq 2$, set $k = v_2(n)$.
If $\Omega_{\operatorname{odd}}(n) = 1$, set $k = v_2(n) - 1$.
If $\Omega_{\operatorname{odd}}(n) = 0$, set $k = v_2(n) - 2$.

In cases 2 and 3, note that $k \geq 0$ since $\Omega(n) \geq 2$. Then set $a = 2^k$ and $b = \tfrac{n}{a}$. Then $n = a \cdot b$, and clearly $a \in A$. We also have $b \in B$:

In case 1 above, $b$ is odd with $\Omega(b) \geq 2$;
In case 2 above, $b$ is even with $\Omega(b) = 2$;
In case 3 above, $b = 4$.

This shows existence of the desired decomposition. For uniqueness, assume $n = a \cdot b$ with $a \in A$ and $b \in B$. Let $m = v_2(n)$. Then $\Omega_{\operatorname{odd}}(b) = \Omega_{\operatorname{odd}}(n)$, so

If $\Omega_{\operatorname{odd}}(n) \geq 2$, then $\Omega_{\operatorname{odd}}(b) \geq 2$, which by definition of $B$ forces $b$ odd, hence $a = 2^m$.
If $\Omega_{\operatorname{odd}}(n) = 1$, then $\Omega_{\operatorname{odd}}(b) = 1$, which by definition of $B$ forces $b$ even and $\Omega(b) = 2$, hence $a = 2^{m-1}$.
If $\Omega_{\operatorname{odd}}(n) = 0$, then $\Omega_{\operatorname{odd}}(b) = 0$, which by definition of $B$ forces $b = 4$, hence $a = 2^{m-2}$.

This shows that $(a,b)$ must be as constructed above. Finally, since all elements of $B$ are composite, any integer of the form $n = a \cdot b$ with $a \in A$ and $b \in B$ is composite. $\square$