A generalisation of Narayana-like numbers (walks on the 2D lattice) I apologize in advance if this question has a trivial answer. I am pretty sure this kind of problem was already studied and I am mostly asking for good references.
Given integers $0 < k \le n+1,$ I am interested in counting strings  $s \in \{-1,1\}^n$ such that 


*

*Every prefix of $s$ has more ones than -1's. In other words the respective sum of the prefix is strictly positive

*For every prefix of $s'$ of $s$ the sum of the elements of $s'$ is less than $k$  


In other words for every prefix $s' = s_1 \cdots s_m$ of $s$ we have $$ 0 < \sum_{i=0}^m s_i < k.$$
Denoting this number by $a(n,k)$ I would like to obtain an asymptotic estimate or upper bound for $a(n,k).$
As far as I can see if I let $k+1$ this reduces to the theory of Dyck words. What am I interested in is the case when $k$ is smaller than $n.$ For example what if $k$ is a positive fraction of $n$? 
 A: Essentially you're asking for the probability of not being absorbed in an unbiased random walk of $n-1$ steps starting at $1$ with absorbing boundaries at $0$ and $k$.  This is a classical topic in probability, and I think e.g. Feller deals with it pretty thoroughly.
EDIT: In the terminology of Feller, chapter XIV, $k = a$, and $u_{1,n}$ is the probability that the process stops at the $n$'th step at barrier $0$, while by symmetry (since $p=q=1/2$) $u_{k-1,n}$ is the probability that it stops at the $n$'th step at barrier $k$.  We then have (from formula (5.7))
$$\eqalign{u_{1,n} &= k^{-1} \sum_{\nu=1}^{k-1} \cos^{n-1}\left( \dfrac{\pi \nu}{k}\right) \sin^2 \left(\dfrac{\pi \nu}{k}\right)\cr
u_{k-1,n} &= k^{-1} \sum_{\nu=1}^{k-1} \cos^{n-1}\left( \dfrac{\pi \nu}{k}\right) \sin \left(\dfrac{\pi \nu}{k}\right) \sin\left(\dfrac{\pi (k-1)\nu}{k}\right)\cr
&=  k^{-1} \sum_{\nu=1}^{k-1} \cos^{n-1}\left( \dfrac{\pi \nu}{k}\right) \sin^2 \left(\dfrac{\pi \nu}{k}\right) (-1)^{\nu-1}\cr
u_{1,n} + u_{k-1,n} &= 2 k^{-1} \sum_{\nu \ \text{odd}} \cos^{n-1}\left( \dfrac{\pi \nu}{k}\right) \sin^2 \left(\dfrac{\pi \nu}{k}\right)}$$
The probability of not yet having been absorbed by the $n$'th step
is 
$$\eqalign{ & \sum_{m=n+1}^\infty (u_{1,m} + u_{k-1,m})\cr
& =  2 k^{-1} \sum_{\nu\ \text{odd}} \dfrac{\cos^n(\pi \nu/k)}{1 - \cos(\pi \nu/k)}\sin^2 \left(\dfrac{\pi \nu}{k}\right)\cr
&=  2 k^{-1} \sum_{\nu \ \text{odd}} \cos^n\left(\dfrac{\pi \nu}{k}\right)\left(1 + \cos\left(\dfrac{\pi \nu}{k}\right)\right) \cr}$$
To get the asymptotics (for $k \to \infty$ with fixed $n$), treat this  as a Riemann sum approximation to
$$ \int_0^1 \cos^n (\pi s) (1 + \cos(\pi s))\; ds = \cases{
 2^{-n} {n \choose n/2} & if $n$ is even\cr
 2^{-n+1} {n-1 \choose (n-1)/2} & if $n$ is odd   } $$
On the other hand, for $n \to \infty$ with fixed $k$, we take the $\nu$ giving the largest
$|\cos(\pi \nu/k)|$, namely $\nu = 1$ (and, if $k$ is even, $\nu = k-1$).
