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Aug 1, 2015 at 14:45 answer added Franz Lemmermeyer timeline score: 2
Aug 1, 2015 at 14:43 history edited Franz Lemmermeyer CC BY-SA 3.0
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Jul 29, 2010 at 14:22 history bounty ended Franz Lemmermeyer
Jul 22, 2010 at 14:00 history bounty started Franz Lemmermeyer
Jul 21, 2010 at 18:22 history edited Franz Lemmermeyer CC BY-SA 2.5
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Mar 21, 2010 at 22:34 comment added Dror Speiser Hmmm. Is this right: I took all monic degree 3 polynomials with coefficients abs value up to 100, with square discriminant that defined different fields. Then computed I the geometric mean of $4 h_K R_k/\sqrt{D_k}$, where K is each such cyclic field. This turned out less than 1. Then I order the fields by discriminant and took partial means, but it remains less than 1. At 100, the value is 0.6546.
Mar 21, 2010 at 21:38 comment added Dror Speiser It does! Lets check for cubics now, and if it works, then this must be what Scholz meant! Also, about the large prime factors condition, is it necessary? The data seemed to suggest it doesn't.
Mar 21, 2010 at 21:06 comment added moonface You should multiply by $\pi$ (that's the constant of proportionality). Then it looks a lot more like $\sqrt{\zeta(2)}$.
Mar 21, 2010 at 19:43 comment added moonface Two things: The density of ideals (as measured by norms) is not the same as the class number; rather, it's proportional to $h(D)/\sqrt{D}$ in the imaginary quadratic case. Secondly, the phrase "with large prime factors" makes a significant difference. I should say I didn't test this experimentally, but playing with it on paper it seems like the answers are as Scholz mentioned.
Mar 21, 2010 at 19:18 comment added Dror Speiser @moonface: I just checked in sage and got: $(\prod_{0 < -D < X} h(D))^{1/r(X)} \sim c\sqrt{X}$ , where $r(X)$ is the number of fundamental negative discriminants greater than $−X$ , and the sum ranges only on those as well. The experimental data says that $c$ is about 0.231, not close to 6. Did I misunderstand?
Mar 21, 2010 at 18:23 comment added moonface ideal density is residue at $1$ of $\zeta_k$; if you take geometric mean of this over fields $k$, I think you get quantities as stated. No idea if this is what he meant.
Mar 21, 2010 at 16:35 answer added Robin Chapman timeline score: 1
Mar 21, 2010 at 15:58 comment added David Loeffler I've retagged this with the generic nt.number-theory tag (more people will see it that way).
Mar 21, 2010 at 15:52 history edited David Loeffler
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Mar 21, 2010 at 15:50 history asked Franz Lemmermeyer CC BY-SA 2.5