How many zero-sum $n$-tuples are there? The question is extremely short and concise.
How many $n$-tuples $X \in \{\, -1,0,1 \,\}^n$ have the zero-sum property
$\sum_{x \in X} x = 0$ ?
At the moment I have nothing to share of my own since I just realized that my previous calculations were wrong.
 A: The number of $+1$ and $-1$ must be the same. So you have to decide:


*

*How many $+1$ (equal to $-1$) are placed, for $\binom{n}{n - 2 k} = \binom{n}{2 k}$ placement options of the $0$

*Think of the $2 k$ places strung out, now you have to decide which ones get $+1$ or $-1$, for $\binom{2 k}{k}$ alternatives.


OK, now stitch all together:
$\begin{align}
\sum_{k \ge 0} \binom{n}{2 k} \binom{2 k}{k}
\end{align}$
maxima claims this isn't Gosper-sumable, so I doubt there is a simpler form.
A: First note that for every $1$ in a tuple we need exactly one $-1$ and the remaining entries must be $0$. So, the multiplicities of $-1,0$ and $1$ in a $n$-tuple are already determined by the number of $1$-entries (we call this number $k$) and $n$.
Now, for some fixed $k$, in how many distinct ways can we rearrange the entries of a $n$-tuple with $k$ $1$-entries? The answer is given by the multinomial coefficients:
$$\binom{n}{k,k,n-2k} = \frac{n!}{k!k!(n-2k)!}$$
Finally, we can have at most $k = \lfloor n/2 \rfloor$ many $1$'s in a $n$-tuple, since any more won't leave enough space for $-1$'s. Therefore, we conclude there are
$$\sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{k,k,n-2k}$$
many $n$-tuples in $\{-1,0,1\}^n$ satisfying the zero-sum property.
(It may be possible to simplify the expression further, but I'm not sure how.)
