Consider an array of (possibly non-distinct) integers $a_0, \ldots , a_{n - 1}$, where $n$ is even, and an additional integer $a$. Suppose that the $a_i$'s $a$-subset sums lack adjacent elements, in the sense that for every subset $\{i_0, \ldots , i_{k - 1}\} \subset \{0, \ldots , n -1 \}$ such that $\sum_{j = 0}^{k - 1} a_{i_j} = a$, it _also_ holds that $\{2 j, 2 j + 1 \} \not \subset \{i_0, \ldots , i_{k - 1}\}$ for each $j \in \{0, \ldots , \frac{n}{2} - 1\}$.

How many subsets $\{i_0, \ldots , i_{k - 1}\} \subset \{0, \ldots , n - 1\}$ can there be such that $\sum_{j = 0}^{k - 1} a_{i_j} = a$? Though the constraint on adjacent elements gives a theoretical maximum of $3^{\frac{n}{2}}$, extensive empirical evidence shows that the actual optimum is $2^{\frac{n}{2}}$. This latter maximum can clearly be attained, for example if $(a_0, a_1, a_2, a_3 \ldots , a_{n - 2}, a_{n - 1}) := (1, 0, 1, 0, \ldots , 1, 0)$ and $a := 0$. Thus the conjecture is that this is the best you can do. I think this should still work if the $a_i$ come from say $\mathbb{Q}$, or any finite field of characteristic $p > n$.

This conjecture also be phrased in terms of boolean and linear algebra. Define the boolean function $f : \{0, 1\}^n \rightarrow \{0, 1\}$, $f(x_0, \ldots , x_{n - 1}) \mapsto (\overline{x_0} \vee \overline{x_1}) \wedge \cdots \wedge (\overline{x_{n - 2}} \vee \overline{x_{n - 1}})$ (this is called the "Achilles heel" function in boolean logic). Then the conjecture is that even though $\left| f^{-1}(1) \right| = 3^{\frac{n}{2}}$, any $\mathbb{Z}$-submodule $A \subset \mathbb{Z}^n$ such that $A \cap \{0, 1\}^n \subset f^{-1}(1)$ actually satisfies $\left| A \cap \{0, 1\}^n \right| \leq 2^{\frac{n}{2}}$, which is well under the _a priori_ maximum.