Consider a binary vector $a_0, a_1,\,\dots\,, a_n$ and an equation
$$\sum_{i=0}^n a_i \cdot (-1)^i {n \choose i} = 0.$$
You can satisfy this trivially when
1) all $a_i$ are 0, or
2) all $a_i$ are 1, or
3) $n$ is odd and $a$ satisfies $a_i = a_{n-i}$ for all $i$, because the summands for $i$ and $n-i$ cancel out.
My question is if there are any other vectors $a$ satisfying the equation?
You are looking for a $n$ dimensional vector $\mathbf{a}$ with components $a_i$ whose Euclidean dot product with the vector $\mathbf{b}$ with components $b_i = {n\choose i}(-1)^i$ is zero. Since the vector $\mathbf{b}$ is fixed and not zero, it generates a 1D subspace of $\mathbb{R}^n$, so there should be an $n-1$ dimensional subspace of vectors which satisfy your condition.
One way to enumerate these naively would be to consider the projection of the standard basis $\mathbf{e}_i$ off your vector $\mathbf{b}$, that is \begin{equation} \mathbf{e}_i' = \mathbf{e}_i - (\mathbf{e}_i\cdot\mathbf{b})\mathbf{b} \end{equation} this will generically form a set of $n$ new vectors $\mathbf{e}'_i$ which are no longer normal or orthogonal and have one linear dependence. You may then perform Gram-Schmidt to find an orthonormal basis which you can use to enumerate this $n-1$ dimensional subspace.
