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Mar 1, 2014 at 19:01 vote accept jamisans
Mar 1, 2014 at 15:40 answer added David E Speyer timeline score: 15
Feb 28, 2014 at 23:24 comment added Eckhard The eigenvectors appear to be integral of small magnitude and containing lots of zeros.
Feb 28, 2014 at 22:35 comment added Jernej It appears that all the graphs are vertex-transitive. If this is indeed so you could try to apply the results from link.springer.com/article/10.1007%2FBF02018821#page-1 or even sciencedirect.com/science/article/pii/0095895679900790
Feb 28, 2014 at 22:17 history edited jamisans CC BY-SA 3.0
Added another example
Feb 28, 2014 at 21:54 answer added Igor Rivin timeline score: 2
Feb 28, 2014 at 21:52 comment added jamisans Perhaps $G_{3,k}$ is the line graph of $K_{k,k,k}$. Also if $n>3$ the (presumed) smallest eigenvalue of $1-n$ would be less than -2, which means that $G_{n,k}$ couldn't be a line graph.
Feb 28, 2014 at 21:41 comment added F. C. sage says that G(3,3), G(3,2), G(4,2), G(4,3) and G(3,4) are not Cartesian products. And also that G(3,3), G(3,4) and G(3,5) are also line graphs.
Feb 28, 2014 at 21:24 answer added F. C. timeline score: 4
Feb 28, 2014 at 21:02 comment added jamisans Thanks. I know at least the displayed graph is not a cartesian product of smaller graphs. It does end up being the line graph of the octahedral graph, however. I will start to look at some other cases to see if they can be decomposed.
Feb 28, 2014 at 20:05 comment added Jernej Can you express them as smaller graphs with a given operation? For example are they ever Cartesian product graphs? Nice construction BTW.
Feb 28, 2014 at 19:52 comment added jamisans I haven't looked at eigenvectors yet - I'll start to do that and see if I can learn anything.
Feb 28, 2014 at 19:38 comment added Alex Degtyarev Do the examples you've done help you guess the eigenvectors?
Feb 28, 2014 at 19:29 history asked jamisans CC BY-SA 3.0