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Nov 4, 2013 at 19:41 vote accept Daniel Hast
Nov 1, 2013 at 21:48 answer added ThisNameForSale timeline score: 8
Nov 1, 2013 at 21:13 comment added Will Sawin The parity conjecture definitely does not force this. We could take the actual set of all elliptic curves and turn a density-$0$ subset of the rank $0$ curves into rank $2$ curves. This would preserve the parity and average rank conjectures, but could change the average rank among curves of rank $\geq 2$, if the set we changed no longer had density $0$ when restricted to that subset of curves.
Nov 1, 2013 at 21:10 answer added Will Sawin timeline score: 11
Nov 1, 2013 at 20:09 answer added Matt Young timeline score: 8
Nov 1, 2013 at 19:47 comment added Dror Speiser I think the parity conjecture forces 2 and 3 to have about the same (once excluding rank 1), and we expect that rank higher than 3 is much more rare. So, one, if one were me, would expect half of elliptic curves of rank at least two to actually be of rank two, another half to be of rank 3, and the rest to have density zero.
Nov 1, 2013 at 19:27 review First posts
Nov 1, 2013 at 19:35
Nov 1, 2013 at 19:08 history asked Daniel Hast CC BY-SA 3.0