Expression of coefficients of a product of Dirichlet polynomials Suppose we have two Dirichlet polynomials:
$$
f_1(s) = \sum_{n=1}^{m} \frac{a_n}{n^s} \\
f_2(s) = \sum_{n=1}^{m} \frac{b_n}{n^s}
$$
Their product will also be a Dirichlet polynomial:
$$
f_1(s)f_2(s) = \sum_{n=1}^{m}\sum_{k=1}^{m} \frac{a_n b_k}{(nk)^s} = \sum_{n=1}^{m^2} \frac{c_n}{n^s}.
$$
Is there a general way to express $c_n$ through $a_n$ and $b_n$? It appears that a variant of Dirichlet convolution can be used, that is,
$$
c_n = \sum_{d:\: d|n,\: d \leq m, \: n/d \leq m} a_d b_{\frac{n}{d}}
$$
if $n \in \{lk : l=\overline{1,m}, k=\overline{1,m}\}$, and $c_n = 0$ otherwise, but such description looks too unwieldy and complex. Is there a simpler way?
 A: Note: There is no simpler way compared to what you have already indicated. But, in fact it's not more complicated than the presumably more familiar Cauchy product of polynomials. Let's make a comparison head-to-head:

We consider two Dirichletpolynomials
  \begin{align*}
f_1(s)=\sum_{n=1}^m\frac{a_n}{n^s} \qquad \text{ and } \qquad f_2(s)=\sum_{n=1}^m\frac{b_n}{n^s}\qquad\qquad (m\geq 1)
\end{align*}
  Multiplication gives
  \begin{align*}
f_1(s)f_2(s)&=\left(\sum_{k=1}^m\frac{a_{k}}{k^s}\right)\left(\sum_{l=1}^m\frac{b_{l}}{l^s}\right)\\
&=\sum_{n=1}^{m^2}\left(\sum_{{1\leq k,l\leq m}\atop{k\cdot l=n}}a_{k}b_{l}\right)\frac{1}{n^s}\\
&=\sum_{n=1}^{m^2}\left(\sum_{{k=1}\atop{k|n}}^{m}a_kb_{\frac{n}{k}}\right)\frac{1}{n^s}
\end{align*}



Now we consider two polynomials of degree $m\geq 1$
  \begin{align*}
h_1(z)=\sum_{n=1}^mc_mz^n \qquad \text{ and } \qquad h_2(z)=\sum_{n=1}^md_mz^m
\end{align*}
  and multiplication gives
  \begin{align*}
h_1(z)h_2(z)&=\left(\sum_{k=1}^mc_kz^{k}\right)\left(\sum_{l=1}^md_lz^{l}\right)\\
&=\sum_{n=1}^{2m}\left(\sum_{{1\leq k,l\leq m}\atop{k+ l=n}}c_kd_l\right)z^n\\
&=\sum_{n=1}^{2m}\left(\sum_{k=1}^{m}c_kd_{n-k}\right)z^n
\end{align*}

