there are large sets $S$ for which $$\label{eqn:size_of_S_k_var}
T(S) {\geqslant}\frac{\sigma_k}{k^{k-1}} |S|^{k-1} + O(|S|^{k-2}).$$

The proof of Proposition \[prop:twelfth\_k\_var\] will rely on the following fact, which states that, when the coefficients $a_i$ are all 1, long progressions behave somewhat like real intervals.

\[prop:intervals\_like\_progressions\] Suppose $S_1, \dots, S_k$ are arithmetic progressions centred at the origin, with the same common difference. Let $s_i$ be the number of terms in $S_i$.

Then, the number of solutions to $x_1 + \dots + x_k = 0$ where each $x_i \in S_i$ is $$\Phi(s_1, \dots, s_k) + O_k((s_1 + \dots + s_k)^{k-2}).$$

We may assume without loss of generality that the progressions $S_i$ have common difference 1. To prove Proposition \[prop:intervals\_like\_progressions\], it suffices to use the following observation.

Suppose that $y_1, \dots, y_{k-1}$ are elements of the real intervals $I_{s_1}, \dots, I_{s_{k-1}}$. Then, we have the following two implications for $k-1$-tuples of real numbers $y_1, \dots, y_{k-1}$.

-   If $|y_1 + \dots + y_{k-1}| {\leqslant}\frac{s_k}{2},$ then $\left|\lfloor y_1\rfloor + \dots + \lfloor y_{k-1}\rfloor \right| {\leqslant}\frac{s_k}{2} + k;$

-   If $\left|\lfloor y_1\rfloor + \dots + \lfloor y_{k-1}\rfloor \right| {\leqslant}\frac{s_k}{2} - k,$ then $|y_1 + \dots + y_{k-1}| {\leqslant}\frac{s_k}{2}.$

Now, $T(S_1, \dots, S_k)$ counts the number of $k-1$-tuples of 