The problem is to evaluate the following sum over all permutations $\sigma\in S_{d}$ of $\{1,2,...,d\}$:

$\displaystyle\sum_{\sigma\in S_{d}}\text{sgn}(\sigma)\displaystyle\frac{1}{\prod_{i=1}^{d}(\sigma_{1} + \sigma_{2} + ... + \sigma_{i})}$.

This is similar to Question 1 here except the sign term. The conjecture is that our sum equals $\prod_{k=1}^{d-1}\frac{(k!)^{2}}{(2k+1)!}$. The problem and the conjecture arises from here.

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    $\begingroup$ Hm, can this perhaps be phrased as a determinant? Smells like some type of Vandermonde-determinant? $\endgroup$ – Per Alexandersson Oct 24 '19 at 10:16
  • $\begingroup$ @PerAlexandersson: That is where it came from: see the original question here. It is the leading coefficient of sum of a Vandermonde polynominal. $\endgroup$ – Abhishek Halder Oct 24 '19 at 13:43
  • $\begingroup$ Here is one thought looking at the conjectured form: $\frac{(k!)^{2}}{(2k+1)!} = \frac{(\Gamma(k+1))^{2}}{\Gamma(2k+2)} = B(k+1,k+1)$. This suggests that the sum may be expressible as product of integrals: $\prod_{k=1}^{d-1}\left(\int_{0}^{1} t_{k}^{k}(1-t_{k})^{k}\:{\rm{d}}t_{k}\right)$. $\endgroup$ – Abhishek Halder Oct 26 '19 at 7:16

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