Multivariate sum over a simplex

Question

Let $s\ge 0$ and $d \ge 0$ be integers. Let $\left\{ a_\eta \right\}_{\eta=0}^s$ be some positive numbers and let $0 \le \xi_1 \le \xi_2 \le \cdots \le \xi_d \le s$. Finally let $k$ be a strictly positive integer. We consider a following multivariate sum: \begin{equation} S^{(s,d)}_{\vec{a},\vec{\xi}} \left(k\right ) := \sum\limits_{1 \le j_0 \le \cdots \le j_s \le k-1}\prod\limits_{\eta=0}^s \binom{j_{\eta+1}-j_\eta-1}{a_\eta} \cdot \prod\limits_{\eta=1}^d j_{\xi_\eta} \end{equation} Clearly when $d=0$ the sum above is a natural generalization of the multivariate beta function to the discrete case. If $d \ge 1$ then the sum is something else yet it is conceivable that it can be related to a mixture of different multivariate beta functions. In view of that we state the following result: \begin{equation} S^{(s,0)}_{\vec{a},\vec{\xi}} \left(k\right ) = \binom{k-1}{A_s+s+1} \end{equation} and \begin{equation} S^{(s,d)}_{\vec{a},\vec{\xi}} \left(k\right ) = \sum\limits_{l=0}^{d-1} \binom{k+l}{A_s+s+2+l} \cdot (-1)^{d-1-l} \cdot (1+\xi_1+A_{\xi_1-1}) \cdot \left\{ \sum\limits_{2\le i_1 < \cdots < i_l \le d} B^{(d)}_l(\vec{i}) \prod\limits_{p=1}^l (p+1+\xi_{i_p}+A_{\xi_{i_p}-1}) \right\} \end{equation} where $A_\xi := \sum\limits_{\eta=0}^\xi a_\eta$ and the coefficients $B^{(d)}_l(\vec{i})$ do not depend on the vector $\vec{a}$. We found this result by using the telescoping property of the binomial factor, meaning just the Pascal triangle identity, and by mathematical induction in the variable $d$. Now the question would be to find those coefficients. Another question would be to find some other way, maybe a combinatorial way, of deriving the result.

Answer 1

Note that the sum in question satisfies a following recursion relation: \begin{equation} S^{(s,d+1)}_{\vec{a},\vec{\xi}}(k) = \sum\limits_{j=1}^{k-1} \binom{k-1-j}{A_s-A_{\xi_{d+1}-1}+s - \xi_{d+1}} \cdot j \cdot S^{(\xi_{d+1}-1,d)}_{\vec{a}}(j) \end{equation} Now, inserting the the conjecture into the recursion relation above we easily get the following recursion relations for the coefficients: \begin{equation} B^{(d+1)}_l(\vec{i}_l) = \left\{ \begin{array}{ll} B^{(d)}_{l-1}(\vec{i}_{l-1} ) & \mbox{if last$(\vec{i}_l)=d+1$}\\ (l+1) \cdot B^{(d)}_l (\vec{i}_l) & \mbox{if last$(\vec{i}_l)\le d$} \end{array} \right. \end{equation} subject to $B^{(d+1)}_0 = 1$. The recursion relation above is easily solved by induction in the variable $l$. We just state the result: \begin{equation} B^{(d+1)}_l\left(i_1,i_2,\cdots,i_l\right) = \prod\limits_{j=2}^{l+1} j^{i_j-1-i_{j-1}} \end{equation} for $2 \le i_1 < \cdots < i_l \le d+1$ subject to $i_{l+1}=d+1$.