Number of $x \in\{-1,1\}^n$ s.t. $ \sum_{j=1}^n x_j=m$ but $\sum_{j=1}^k x_j\neq m$ for all $k \in\{1,\ldots,n-1\}$ I came across this combinatorial problem:

Let $m,n\in {\mathbb N}$, then what is the number of $x \in\{-1,1\}^n$ such that $ \sum_{j=1}^n x_j=m$ but $\sum_{j=1}^k x_j\neq m$ for all $k \in\{1,\ldots,n-1\}$?

There $x=(x_1,\ldots ,x_n)$. In reality Im not interested in the exact numbers, Im more interested in some good approximation when $n\gg m $. My work so far: set such numbers as $c_{m,n}$, then
$$
\binom{n}{\frac{m+n}{2}}= \sum_{j\geqslant 0}c_{m,n-j}\binom{j}{\frac{j}2}\quad \text{  and }\quad c_{m+2,n+2}=c_{m+2,n}+c_{m,n} 
$$
where $ \binom{n}{k}=0$ if $k \notin {\mathbb N}$ and $c_{m,n}=0$ when $m>n$. However after playing some time with these identities I dont found something useful, so I ask here for some help.

UPDATE: trying to solve the recurrence I get
$$
H_{m,n}=\frac{(a_{0,\bullet }+a_{1,\bullet })(1-y^2)-xy}{1-y^2-x^2y^2}\\
\text{ for }\quad H_{m,n}:= \sum_{m,n\geqslant 0}c_{m,n}x^my^n \quad \text{ and }\quad a_{j,\bullet }:= \sum_{n\geqslant 0}c_{j,n}x^jy^n
$$
but I dont get something useful from here.
 A: Removing the last term of the sequence (which must always be $1$), let us think of $c_{m,n}$ as the number of $x\in\{-1,1\}^{n-1}$ such that the sum is $m-1$ and no partial sum goes above $m-1$.  This makes it clear that these are a generalization of Catalan numbers; indeed, $c_{1,2n+1}$ is just the $n$th Catalan number $\frac{1}{n+1}\binom{2n}{n}$.
We can compute $c_{m,n}$ in general by a similar trick as is used to obtain the closed form for Catalan numbers (as in https://en.wikipedia.org/wiki/Catalan_number#Second_proof, for instance).  Namely, given $x\in\{-1,1\}^{n-1}$ whose sum is $m-1$ and such that some partial sum is greater than $m-1$, negate all the terms of $x$ after the first such partial sum.  This defines a bijection between the set of all such $x$ and the set of all $y\in \{-1,1\}^{n-1}$ whose sum is $m+1$.  The number of such $x$ is $$\binom{n-1}{\frac{m+n}{2}-1}-c_{m,n}$$ (the first term counts all sequences of length $n-1$ whose sum is $m-1$) and the number of such $y$ is $$\binom{n-1}{\frac{m+n}{2}}.$$
Thus $$c_{m,n}=\binom{n-1}{\frac{m+n}{2}-1}-\binom{n-1}{\frac{m+n}{2}}=\frac{2m}{m+n}\binom{n-1}{\frac{m+n}{2}-1}.$$
