Restricted Sum Simplification

Question

I am trying to find the coefficients $b_i$ for a restrictive sum that I am trying to simplify: $$\sum_{\sum_{k=1}^K x_k=X} e^{\alpha \sum_{k=1}^K kx_k}=\sum_{i=X}^{XK}b_ie^{\alpha i}$$ (where $x_k$ are positive integers) and scoured a bit trying to find some identities, but the closest I found was the multinomial theorem. https://en.wikipedia.org/wiki/Multinomial_theorem

Was wondering if anyone knew of any references/resources that deal with restrictive sums.

UPDATE: I recently found that a way to generate this polynomial is via q-Binomials (also known as Gaussian binomials https://en.wikipedia.org/wiki/Gaussian_binomial_coefficient) via: $$ \binom{X+K-1}{X}_q q^X=\sum_{i=X}^{XK}b_ie^{\alpha i}$$ where $q=e^\alpha$. However there still does not seem to be a closed form for $b_i$.

Answer 1

Here's a calculation of the generating function. If you call your sum $a_X$, you seem to be asking for $f(z)=\sum_X\, a_X\,z^X$. This function is the product of $K$ generating functions $f_k(z)$ for $k=1..K$, where each $$f_k(z)=\sum_{\ell\ge1}e^{\alpha k \ell}z^\ell=\frac{e^{\alpha k}z}{1-e^{\alpha k }z}.$$ Multiplying these, we get $$f(z)=\prod_{k=1}^K \frac{e^{\alpha k}z}{1-e^{\alpha k }z}=\frac{z^Ke^{\alpha K(K+1)/2}}{\prod_{k=1}^K (1-e^{\alpha k }z)}.$$ In just about any application, you'll have to deal with the singularities at $z=e^{-\alpha k}$, but that seems to be a different question.