Writing a sum over a square lattice as sum over 1D chains in a circle
Consider the sum
\begin{equation}
S = \sum_{m=1}^\infty \sum_{n=1}^\infty V(m,n).
\end{equation}
Here we are summing over a square grid of points $\left\{(m,n)|m,n \in \mathbb{N}\right\}$. Alternatively, these points can be constructed from many 1D chains at different angles. Each chain can be parameterised by the integers $a$ and $b$, and consists of the points $\left\{(na,nb)|n \in \mathbb{N}\right\}$. To reconstruct the square grid without double counting, we must restrict $a$ and $b$ to be coprime. Therefore we consider the set of coprime natural numbers $C=\left\{ (a,b) | a,b \in \mathbb{N}\text{, and } a \perp b \right\}$. In this way we can rewrite the sum as
\begin{equation}
S = \sum_{(a,b)\in C} L(a,b),
\end{equation}
where
\begin{equation}
L(a,b) = \sum_{n=1}^\infty V(n a, n b).
\end{equation}
Let's try $V(m,n)=\frac{1}{(m^2+n^2)^2}$ as an example
As a first example let's consider an 'interaction' term $V(m,n)=\frac{1}{(m^2+n^2)^2}$. Just as a point of interest, we note that the sum exactly evaluates to $S=K\zeta(2)-\zeta(4)$ where $K$ is the Catalan number and $\zeta$ is the Riemann Zeta function [ref]. In addition the 1D sum evaluates to
\begin{equation}
L(a,b) = \frac{\zeta(4)}{(a^2 + b^2)^2}.
\end{equation}
Therefore the full sum can be written as
\begin{equation}
S = \zeta(4) \sum_{(a,b)\in C} \frac{1}{(a^2+b^2)^2}.
\end{equation}
Is this useful? Who knows. Numerically, both sums execute and converge identically. But maybe there are some clever algorithmic things that can be done with the 'circle sum' to speed it up.
The originally proposed summand $V_p(m,n)=\frac{\cos(m k_x + n k_y)}{(m^2 + n^2)^{p/2}}$
Now let us consider the 'interaction' term shown in the subsection title. The 1D version of this summed is the Clausen function
\begin{equation}
\text{Cl}_p(k) = \sum_{n=1}^\infty \frac{\cos(nk)}{n^p}.
\end{equation}
Therefore the 1D sum evaluates to
\begin{equation}
L_p(a,b) = \frac{\text{Cl}_p(a k_x + b k_y)}{\left( a^2 + b^2 \right)^{p/2}},
\end{equation}
and we can write the circle sum as
\begin{equation}
S_p(k_x,k_y) = \sum_{(a,b)\in C} \frac{\text{Cl}_p(a k_x + b k_y)}{\left( a^2 + b^2 \right)^{p/2}}.
\end{equation}
