Formula for $P(d) = \sum_{\sum_{i=1}^m k_i = d, k_i \in \mathbb{N}_+} \left( \prod_{i=1}^m k_i^{a_i} \right) $

Question

Let $\mathbb{N}_+ = \{ 1, 2, \dots\} $. For a given sequence of elements $\{a_i \}_{1 \leq i \leq m} $in $ \mathbb{N}_+ $, we define

\begin{equation} P(d) = \sum_{\sum_{i=1}^m k_i = d, k_i \in \mathbb{N}_+} \left( \prod_{i=1}^m k_i^{a_i} \right). \end{equation}

I know that $ P(d)$ is a polynomial in $d$, and that $\deg(P(d) ) = \sum_{i=1}^m a_i + m - 1 $, but I would like to know if there is a more explicit expression for it.

At the very least, I suspect that the leading coefficient should be computable, but I have not found relevant formulas, as I am unsure about the terminology for this type of sum.

Answer 1

Fix $m$, and $a_1,\ldots,a_m$. Let $\mathbf{x}=(x_1,\ldots,x_m)$ and $\mathbf{y}=(y^1_1,\ldots,y^1_{a_1},\ldots,y^m_1,\ldots,y^m_{a_m})$. Consider the set of points $\mathcal{P}\subseteq \mathbb{R}^{m+\sum a_i}$ defined by $$\mathcal{P}=\{(\mathbf{x},\mathbf{y})\colon x_i \geq 0, \, y^i_j \geq 0, \, y^i_j \leq x_i, \, x_1+\cdots+x_m=1\}.$$ Then $\mathcal{P}$ is a $(m-1+\sum a_i)$-dimensional convex polytope in $\mathbb{R}^{m+\sum a_i}$. A priori $\mathcal{P}$ is just a rational polytope, but I think it should be possible to show that its vertices are in fact lattice points - they will be 0-1 vectors.

Your $P(d)$ is its Ehrhart polynomial of $\mathcal{P}$. More precisely, your $P(d+m)$ is $L(\mathcal{P},d)$, since you require all $k_i \geq 1$. This means that the leading coefficient of $P(d)$ is the (normalized) volume of $\mathcal{P}$. Of course, whether this is helpful for you depends on what you want to do.