Non-trivial alternating sums of binomial coefficients 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?
[There has been a related question but regarding $a_i \in \{-1,1\}$. There are non-trivial examples given in the answers but they do not seem to work in this setup. I have also asked this at math.stackexchange but then I read that this community if better suited for research-level questions.]
 A: What you have is the $n$-th difference operator applied to the sequence $(a_i)$.
In particular, the value is 0 if $a_i=f(i)$ for any polynomial $f$ of degree less than $n$.
The converse is also true.  If the sum is 0 it means that the $n$-th difference is 0, which means there is a polynomial $f$ of degree less than $n$ such that $a_i=f(i)$ for all $i$.  This is therefore a complete characterisation. 
A: ${8\choose{2}}-{8\choose{5}}+{8\choose{6}}=28-56+28=0$, so there's a solution for $n=8$.
A: The binomial coefficients ${n\choose k-1}, {n\choose k}$, and ${n\choose k+1}$ are in arithmetic progression if and only if $n=m^2-2$ and $k=\frac 12(n-m)$ or $\frac 12(n+m)$. For such $n$ and $k$ we have
     $$ {n\choose k-1}-{n\choose k}+{n\choose k+1}-{n\choose n-k}=0. $$
A: (Reposting my comment as an answer, as requested by the OP.)
If you have a solution with $a_i \in \{−1,1\}$, then you also have a solution with $a_i \in \{0,1\}$ simply by replacing each $a_i$ with $(a_i+1)/2$ (using the fact that setting all $a_i=1$ is also a valid solution). In other words, your question had already been answered by some of the comments to this question.
