Riemann vs 2018

For two functions $f,g$ with domain $\mathbb{N}$ , the Dirichlet convolution $f \star g$ is the function $(f \star g)(n) \; = \; \sum_{d|n} f(d) g\big(\tfrac{n}{d}\big) \hspace{2cm} n \in \mathbb{N}$ For any function $f$ with domain $\mathbb{N}$ define $S_f(s) \; = \; \sum_{n \ge 1} \frac{f(n)}{n^s}$ for $s > 0$ large enough to ensure convergence (if possible). It is a standard piece of bookwork that if $S_f(s)$ and $S_g(s)$ are absolutely convergent for $s > T$ , then $S_{f \star g}(s)$ is also absolutely convergent for $s > T$ , and that $S_{f \star g}(s) \; = \; S_f(s) S_g(s) \hspace{2cm} s > T$ Note that $\psi = 1 \star 1$ and $\sigma = 1 \star i$ where $1(n) = 1$ , $i(n) = n$ for all $n \ge 1$ . Since $S_1(s) = \zeta(s)$ for $s > 1$ and $S_i(s) = \zeta(s-1)$ for $s > 2$ , we deduce that $\begin{array}{rclll} S_\psi(s) &=& \zeta(s)^2 & \hspace{1cm} & s > 1 \\ S_\sigma(s) &=& \zeta(s)\zeta(s-1)& & s > 2 \end{array}$

Thus we have $A = B = 2018$ , making the answer $\boxed{1009}$ .