Problem
We have the recurrence relation
$$a(j, k, n) = \begin{cases} 1 & : k = 0 ∧ j = n − 1 \\ 0 & : k = 0 ∧ j < n − 1 \\ a(j, k − 1, n) + (−1)^{k + n − 1 − j} × \frac{n!} {k! (n − k)!} × \frac{(n − 1)!} {j! (n − 1 − j)!} × k^{n − 1 − j} & : k > 0 \end{cases}$$
Can this be converted to an explicit equation in terms of $j$, $k$, and $n$?
Where the Relation Comes From
The recurrence relation generates coefficients for the piecewise equation of the Probability Density Function (PDF) of an Irwin-Hall Distribution (IHD), the sum of $n$ uniformly-distributed, independent, & random variables, over the interval $k ≤ x ≤ k + 1$. These coefficients are plugged into the expression
$$f_X(x;n) = \frac{1}{(n − 1)!} × \sum_{j = 0}^{n − 1} \left(a(j, k, n) × x^j\right)$$
to obtain the full equation.
My Work So Far
It is known that the solution to a recurrence relation in the form $a(k) = a(k - 1) + f(k)$ is equal to $a(k) - a(0) = \sum_{m = 1}^k f(m)$. Thus, if $\sum_{m = 1}^k f(m)$ has a closed formula, you're good to go.
This recurrence relation does seem to be in that form, but I haven't been able to find a closed-form expression for $\sum_{m = 1}^k \left( (−1)^{m + n − 1 − j} × \frac{n!} {m! (n − m)!} × \frac{(n − 1)!} {j! (n − 1 − j)!} × m^{n − 1 − j} \right)$. I'm not sure if the presence of two additional variables makes this method invalid or just more complicated.
Evaluating that sum gives the infinite series
$$ \frac{n! × (n − 1)!} {j! × (n − 1 − j)!} × \frac{(−1)^{n − j + 0} × 1^{n − 1 − j}} {1! × (n − 1)!} + \\ \frac{n! × (n − 1)!} {j! × (n − 1 − j)!} × \frac{(−1)^{n − j + 1} × 2^{n − 1 − j}} {2! × (n − 2)!} + \\ \frac{n! × (n − 1)!} {j! × (n − 1 − j)!} × \frac{(−1)^{n − j + 2} × 3^{n − 1 − j}} {3! × (n − 3)!} + \\ \frac{n! × (n − 1)!} {j! × (n − 1 − j)!} × \frac{(−1)^{n − j + 3} × 4^{n − 1 − j}} {4! × (n − 4)!} + \\ \frac{n! × (n − 1)!} {j! × (n − 1 − j)!} × \frac{(−1)^{n − j + 4} × 5^{n − 1 − j}} {5! × (n − 5)!} + \\ \frac{n! × (n − 1)!} {j! × (n − 1 − j)!} × \frac{(−1)^{n − j + 5} × 6^{n − 1 − j}} {6! × (n − 6)!} + \\ … $$
The partial sums of this infinite series are
- $\frac{n! × (n − 1)! × (−1)^{n − j}} {j! × (n − 1 − j)!} × \left(\frac{1^{n − 2 − j}} {0! × (n − 1)!}\right)$
- $\frac{n! × (n − 1)! × (−1)^{n − j}} {j! × (n − 1 − j)!} × \left(\frac{1^{n − 2 − j}} {0! × (n − 1)!} − \frac{2^{n − 2 − j}} {1! × (n − 2)!}\right)$
- $\frac{n! × (n − 1)! × (−1)^{n − j}} {j! × (n − 1 − j)!} × \left(\frac{1^{n − 2 − j}} {0! × (n − 1)!} − \frac{2^{n − 2 − j}} {1! × (n − 2)!} + \frac{3^{n − 2 − j}} {2! × (n − 3)!}\right)$
- $\frac{n! × (n − 1)! × (−1)^{n − j}} {j! × (n − 1 − j)!} × \left(\frac{1^{n − 2 − j}} {0! × (n − 1)!} − \frac{2^{n − 2 − j}} {1! × (n − 2)!} + \frac{3^{n − 2 − j}} {2! × (n − 3)!} − \frac{4^{n − 2 − j}} {3! × (n − 4)!}\right)$
- $\frac{n! × (n − 1)! × (−1)^{n − j}} {j! × (n − 1 − j)!} × \left(\frac{1^{n − 2 − j}} {0! × (n − 1)!} − \frac{2^{n − 2 − j}} {1! × (n − 2)!} + \frac{3^{n − 2 − j}} {2! × (n − 3)!} − \frac{4^{n − 2 − j}} {3! × (n − 4)!} + \frac{5^{n − 2 − j}} {4! × (n − 5)!}\right)$
- $…$
This pattern doesn't seem to collapse into a nice closed form, so it may be a dead end, but I don't know how to be sure.