A special Vandermonde system with integer coefficients

Question

Consider the Vandermonde system $$\left(\begin{matrix} 1&1&1&\dots&\dots&1\\ 1&2&3&4&\dots&n\\ 1&2^2&3^2&4^2&\dots&n^{2}\\ 1&2^3&3^3&4^3&\dots&n^{3}\\ \vdots&&&&&\vdots\\ 1&2^{n-1}&3^{n-1}&4^{n-1}&\dots&n^{n-1}\\ \end{matrix} \right)~ \left(\begin{matrix} x_{1,n}\\ x_{2,n}\\\dots\\\vdots\\ \dots \\ x_{n,n} \end{matrix}\right) =\left(\begin{matrix} 1\\-1\\1\\-1\\ \vdots\\ (-1)^{n-1} \end{matrix} \right) $$ The solution for $n=6$ is $(21, -70, 105, -84, 35, -6)$. Is there a formula for the general solution involving 2 parameters $j$ and $n $ and, may be, binomial coefficients?

Answer 1

With help from OEIS A127717 we conjecture that for $0\le p\lt n$

$$\sum_{k=1}^n k^p (-1)^{k+1} \sum_{q=k}^n {q-1\choose k-1} q = (-1)^p.$$

This is

$$\sum_{q=1}^n q \sum_{k=1}^q {q-1\choose k-1} (-1)^{k+1} k^p = \sum_{q=1}^n q \sum_{k=0}^{q-1} {q-1\choose k} (-1)^{k} (k+1)^p \\ = \sum_{q=1}^n q \sum_{k=0}^{q-1} {q-1\choose k} (-1)^{k} p! [z^p] \exp((k+1)z) \\ = p! [z^p] \exp(z) \sum_{q=1}^n q \sum_{k=0}^{q-1} {q-1\choose k} (-1)^{k} \exp(kz) \\ = p! [z^p] \exp(z) \sum_{q=1}^n q (1-\exp(z))^{q-1}.$$

Now with $(1-\exp(z))^{q-1} = (-1)^{q-1} z^{q-1} + \cdots$ and $n\gt p$ we may certainly extend $q$ to infinity without contributing to $[z^p]$ and we find

$$p! [z^p] \exp(z) \sum_{q\ge 1} q (1-\exp(z))^{q-1} \\ = p! [z^p] \exp(z) \frac{1}{(1-(1-\exp(z)))^2} \\ = p! [z^p] \exp(z) \exp(-2z) = p! [z^p] \exp(-z) = (-1)^p.$$

This is the claim. We see that the desired coefficient is given by

$$\bbox[5px,border:2px solid #00A000]{ x_{k,n} = (-1)^{k+1} \sum_{q=k}^n {q-1\choose k-1} q.}$$

Remark. A careful examination of the OEIS entry shows that we can simplify $x_{k,n}.$ We get for the sum

$$\sum_{q=k}^n {q-1\choose k-1} q = k \sum_{q=k}^n {q\choose k} = k \sum_{q=0}^{n-k} {q+k\choose k} \\ = k \sum_{q=0}^{n-k} [z^k] (1+z)^{q+k} = k [z^k] (1+z)^k \sum_{q=0}^{n-k} (1+z)^q \\ = k [z^k] (1+z)^k \frac{(1+z)^{n-k+1}-1}{1+z-1} \\ = k [z^{k+1}] (1+z)^k ((1+z)^{n-k+1}-1) \\ = k [z^{k+1}] (1+z)^k (1+z)^{n-k+1} = k [z^{k+1}] (1+z)^{n+1}.$$

We thus have

$$\bbox[5px,border:2px solid #00A000]{ x_{k,n} = (-1)^{k+1} k {n+1\choose k+1}.}$$