Skip to main content
2 events
when toggle format what by license comment
Oct 4, 2020 at 21:26 comment added Sam Hopkins Note that if we drop the "with 0 diagonal" condition, the answer is known exactly. Namely, the Ehrhart quasi-polynomial then is of the form $P_n(r)+(-1)^rQ_n(r)$ where the degree of $P_n(r)$ is $\binom{n}{2}$ and the degree of $Q_n(r)$ is $\binom{n-1}{2}-1$ if $n$ is odd and $\binom{n-2}{2}-1$ if $n$ is even. See Stanley, "Magic labelings of graphs, symmetric magic squares ..." (doi.org/10.1215/S0012-7094-76-04342-8) and Jia, "Symmetric magic squares and multivariate splines" (doi.org/10.1016/0024-3795(95)00451-3).
May 24, 2018 at 20:30 history answered Brendan McKay CC BY-SA 4.0