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Timeline for Hamiltonian cycles in power-graphs

Current License: CC BY-SA 3.0

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Jan 17, 2013 at 17:32 comment added Felix Goldberg @Ami Paz: I thought about something like this, but doesn't the finite field make a huge difference? Maybe there is a version of Paley graphs over R?
Jan 17, 2013 at 14:11 comment added Ami Paz One must wonder if this has to do with Paley graphs: Paley graph is defined similarly, but over a finite field, and there is a edge iff the difference is a square in the field. They are known to be Hamiltonian.
Jan 17, 2013 at 13:34 answer added Gerhard Paseman timeline score: 1
Jan 17, 2013 at 12:52 comment added Jernej That's a very nice question! I have tested the conjecture for values of $n$ up to 500 and it holds.
Jan 17, 2013 at 9:54 history edited Felix Goldberg CC BY-SA 3.0
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Jan 17, 2013 at 9:51 comment added Felix Goldberg @Olivier: I am also puzzled by Gerhard's assertion; however, do note that the numbers begin to begin monotonically from 37. The OEIS has terms up to 52, and they all increase (oeis.org/A071984).
Jan 17, 2013 at 9:40 comment added Olivier Dear Gerhard, Can you explain why you think this implication is reasonable? I don't see it myself (but I don't know much about the topic). Moreover, the data in the linked note show that there exists values of $n$ such that G2(n+2) has less hamiltonian cycles than G2(n).
Jan 16, 2013 at 22:01 comment added Felix Goldberg @verret: Yes, of course! Thanks for spotting the error. I corrected the question.
Jan 16, 2013 at 22:00 history edited Felix Goldberg CC BY-SA 3.0
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Jan 16, 2013 at 17:49 comment added verret Do you mean $G_2(32)$ instead of $G_{32}(2)$?
Jan 16, 2013 at 17:16 comment added Gerhard Paseman It seems reasonable that G2(n) is Hamiltonian would imply G2(n+2) also is. I don't know how else to attack it. Gerhard "Ask Me About System Design" Paseman, 2013.01.16
Jan 16, 2013 at 12:45 history asked Felix Goldberg CC BY-SA 3.0