Solve the recurrence \[ f_{j,k}^{(l)} = \begin{cases} \left[j>k\right] j^{k-1}(j-k), &\qquad j=l \\ \\ \left[j>k+1\right] \sum_t \binom k t f_{j-1,k-t}^{(l)}, &\qquad j>l \end{cases} \] for all nonnegative integer $k$ and postive integers $j$, $l$, where
\[ [P] = \begin{cases} 1, &\qquad \hbox{if $P$ is true} \\ 0, &\qquad \hbox{otherwise} \end{cases} \]
is Iverson bracket, see Wikipedia. For example, $[2 < 3] = 1; [2 > 3] = 0$
The problem is introduced from a programming problem from URAL. I'm trying to find a closed form solution for the problem (notice that $f_{j,k}^{(l)}$ is the number of arrangements of $j$ horses and $k$ students in which the last horse is empty and the adjacent $l$ horses are not all occupied).
Here I want to show an algebraic way, to obtain $f_{j,k}^{(l)} = [j>k] j^{k-1}(j-k), \qquad j=l$ which I got through combinatorial interpretation, that I think helpful to solve the recurrence of $f_{j,k}^{(l)}$. In fact, $[j>k] j^{k-1}(j-k)$ is the solution to the following recurrence:
\begin{align*} g_{1,k} &= \left[k \le 1\right] (1 - k), \qquad k \ge 0 \\ g_{j,k} &= \left[k < j\right] \sum_t \binom k t g_{j-1,k-t}, \qquad j > 1, k \ge 0 \end{align*}
Notice that the recurrence of $g$ is similiar with that of $f^{(l)}$. It seems to be as strange as the recurrence of $f^{(l)}$, but it isn't too messy.
Let's solve it by hand. First assume that
\begin{align*} g_{1,k}^* &= 1 - k, \qquad k \ge 0 \\ g_{j,k}^* &= \sum_t \binom k t g_{j-1,k-t}, \qquad j > 1, k \ge 0 \end{align*}
We have $g_{1,k}^* = g_{1,k}$ for $0 \le k \le 1$. Now let's consider the exponential generating function for $g_{j,k}^*$: $\hat G_j^*(z) = \sum_{k \ge 0} g_{j,k}^* z^k/k!$. \begin{align*} \hat G_1^*(z) &= \sum_{k \ge 0} \frac{(1-k)z^k}{k!} \\ &= \sum_{k \ge 0} \left(\frac{z^k}{k!} - \frac{kz^k}{k!}\right) \\ &= \sum_{k \ge 0} \frac{z^k}{k!} - \sum_{k \ge 0} \frac{z^{k+1}}{k!} \\ &= (1-z)e^z \end{align*} and for $j > 1$, \begin{align*} \hat G_j^*(z) &= \sum_{k \ge 0} \sum_{0 \le t \le k} \frac{g_{j-1,k-t}^*z^k}{t!(k-t)!} \\ &= \sum_{t \ge 0} \frac{z^t}{t!} \sum_{k \ge t} \frac{g_{j-1,k-t}^*z^{k-t}}{(k-t)!} \\ &= \sum_{t \ge 0} \frac{z^t}{t!} \sum_{k \ge 0} \frac{g_{j-1,k}^*z^k}{k!} \\ &= e^z \hat G_{j-1}^*(z) \end{align*} so $\hat G_j^*(z) = (1-z)e^{jz}$, and we get $g_{j,k}^* = j^{k-1}(j-k)$, and $g_{k,k}^* = g_{k,k} = 0$ whenever $k \ge 1$.
Now it's easy to prove $g_{j,k} = g_{j,k}^*$ for $0 \le k \le j$ by induction on $j$.
Can we solve $f_{j,k}^{(l)}$ like this? More precisely, extend some zeros of $f_{j,k}^{(l)}$ to nonzeros to eliminate the awful factor $[j > k + 1]$ in recurrence, solve it and prove that the nonzeros of $f_{j,k}^{(l)}$ keeps its value when extending?
Thanks for your help!