I'm trying to find a closed form expression to the following recurrence relation:
\begin{align} &a_k(0) = 0, \quad \forall k\geqslant 1; \\ &a_k(1) = 1, \quad \forall k\geqslant 1; \\ &a_0(n) = 1, \quad \forall n\geqslant 1; \\ &a_k(n) = \sum_{0\leqslant j<n} a_{k-1}(n+j). \end{align}
The case $k=2$ are particularly interesting, since it gives the pentagonal numbers,
$$ a_2(n) = \sum_{0\leqslant j<n} a_1(n+j) = \sum_{0\leqslant j<n} (n+j) = \binom{2n}{2} - \binom{n}{2} = \frac{n(3n-1)}{2}.$$
But for higher $k$ it seems to behave very "chaotically" (in a not exactly strict sense of this word).
If instead of "$n+j$" one chooses "$n-j$" on the definition, the closed form of the recurrence becomes much more friendly, namely $\binom{n+k-1}{k}$, that is well-known counting way, and even can be extended to the complex function in two variables $\frac{\Gamma(z+w)}{\Gamma(z)\Gamma(w+1)}$. It makes me wonder whether the above recurrence relation also have a combinatorial interest, or even a closed form that can be naturally extended to complex variables.