A polynomial power-compositions determinant

Josep M. Brunat¹

Departament de Matemàtica Aplicada II
Universitat Politècnica de Catalunya
C. Pau Gargallo, 5. E-08028 Barcelona, Catalonia, Spain
Josep.M.Brunat@upc.edu

Let $n$ and $p$ be positive integers. A $p$-composition of $n$ is a $p$-tuple of non-negative integers $\alpha=(\alpha_1,\ldots,\alpha_p)$ such that $\alpha_1+\cdots+\alpha_p=n$. Denote by $C(n,p)$ the set of $p$-compositions of $n$. If $\alpha=(\alpha_1,\ldots,\alpha_p)$ and $\beta=(\beta_1,\ldots,\beta_p)$ are $p$-compositions of $n$, we denote $\alpha^\beta={\alpha_1}^{\beta_1} \cdots {\alpha_p}^{\beta^p},$ where to be consistent, it is assumed that $0^0=1$. The power-compositions determinant is the determinant

$$\Delta(n,p)=\det_{\alpha,\beta\in C(n,p)} (\alpha^\beta).$$

The value of $\Delta(n,p)$ is given in [1]:

$$\Delta(n,p)=\prod_{k=1}^{\min\{n,p\}}\left( n^{n-1 \choose... ...n-k+1} i^{(n-i+1){n-i-1 \choose k-2}}\right)^{p \choose k}.$$

Recently, C. Krattenthaler in the complement [3] to its impressive Advanced Determinant Calculus[2], has given an equivalent formula for $\Delta(n,p)$ and has stated the following conjecture supported by computer experiments:

$$\det_{\alpha,\beta\in C(n,p)} ((x+\alpha)^\beta) =(px+n)^{n+p-1\choose p}\prod_{i=1}^n i^{(n-i+1){n+p-i-1\choose p-2}},$$

where $x$ is a variable and $x+\alpha$ is short for $(x+\alpha_1,\ldots,x+\alpha_p)$. In this talk we prove this conjecture using a method that can be useful for other combinatorial determinants.

Footnotes

... Brunat¹

joint work with Antonio Montes