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Jun 16, 2010 at 21:24 comment added Wadim Zudilin Martin, thank you for the tip. No, I didn't. But I'll try to do so today.
Jun 16, 2010 at 16:08 comment added Martin Rubey Did you try to interpret the determinant as generating function for non-intersecting lattice paths via Stembridge's theorem (Advances in Mathematics 83, Nonintersecting Paths, Pfaffians, and Plane Partitions)?
Jun 16, 2010 at 14:25 comment added Wadim Zudilin Victor and Robin, thanks for notifying this mistake. Yes, of course, I meant that only the squares have to be taken in the range.
Jun 16, 2010 at 14:18 comment added Victor Protsak Wadim, Robin was commenting on your statement that "det(V) seems to assume all possible odd positive values restricted only by Hadamard's_inequality": there is an additional requirement that det(V) be a perfect square (or you should say Pfaffian).
Jun 16, 2010 at 9:09 comment added Wadim Zudilin Robin, I provide the required links which explain that $\det(V)$ is the square of the Pfaffian. I can even compute the Pfaffian as a single determinant of a $(n/2)\times(n/2)$ matrix in the case $v_{ij}=\epsilon_{i-j}$. A tournament is another way to say a complete signed (oriented) graph, but I am looking for a structure of graphs satisfying $\det(V)=1$.
Jun 16, 2010 at 8:28 comment added Robin Chapman This combinatorial structure (a complete graph with an orientation) is known as a tournament. Also, as a skew-symmetric matrix, $\det(V)$ is a square (the square of the Pfaffian of $V$).
Jun 16, 2010 at 8:08 history asked Wadim Zudilin CC BY-SA 2.5