most recent 30 from http://mathoverflow.net 2013-05-20T07:00:39Z http://mathoverflow.net/feeds/user/1936 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111173/how-long-can-a-primal-egyptian-fraction-be-that-optimally-approaches-unity/111184#111184 Stefano Pascolutti 2012-11-01T17:37:02Z 2012-11-01T17:37:02Z

<p>$$\frac{1}{2} + \frac{1}{3} + \frac{1}{11} + \frac{1}{23} + \frac{1}{31} + \frac{1}{2\cdot 3\cdot 11\cdot 23\cdot 31+1} = 1 - \frac{1}{2214502422}$$ and $2\cdot 3\cdot 11\cdot 23\cdot 31+1$ is prime (and 2214502422 is the product of the denominators).</p>

http://mathoverflow.net/questions/111132/possible-slope-for-a-modular-form Stefano Pascolutti 2012-11-01T10:30:10Z 2012-11-01T10:30:10Z

<p>Let $f\colon \mathbb{H}_g \to \mathbb{C}$ be a Siegel modular form of weight $k$ with respect to $\Gamma_g$. Then, $f$ admits a Fourier expansion $f(Z) = \sum_T a(T) \exp(i\pi \mathop{tr} TZ)$, where $T$ varies amongst all even symmetrical semipositive definite matrices. The number $v := \frac{1}{2}\min_{x \in \mathbb{Z}^g \setminus 0} \{x^tTx \mid a(T) \neq 0\}$ is the <em>vanishing order</em> of $f$. A <em>cusp</em> form is a modular form with $v > 0$. (Check, for example, Slope of Integral Lattices, C.Poor, D.Yuen)</p> <p>We define the slope of a Modular cusp form $\mathop{sl}(f) = \frac{k}{v}$, which is a rational positive number.</p> <blockquote> <p>Is there a modular cusp form with slope $q$ for every rational positive $q$?</p> </blockquote> <p>If $g$ is fixed, I think the answer is false (this is related to the minimale slope of $\mathcal{A}_g$, which is greater than 0 -and it is only 0 asymptotically). So a better questions should be the following:</p> <blockquote> <p>If $g$ is fixed, is there a modular cusp form with slope $q$ for every rational number $q > \bar{q}$, for some $\bar{q} = \bar{q}(g) \in \mathbb{Q}$?</p> </blockquote> <p>If the answer is known to be negative, how can I possibly prove that there is no Modular cusp form with slope $q$ for a $q$ which contradicts the claim? If the answer is known to be positive, how can I produce a modular form with slope $q$?</p> <p>The idea behind this is to use modular forms to define slope of loci inside $\mathcal{A}_g$. For example, we have that the slope of the hyperelliptic locus is $8 + \frac{4}{g}$ since any modular form with slope $&lt; 8 + 4/g$ vanishes on the hyperelliptic locus and this is sharp (we have a modular form with slope $8 + 4/g$ which does not vanish on the hyperelliptic locus -see R. Salvati-Manni "Slope of cusp forms and Theta series" JNT (2000)). The problem is that in general, modular forms could not be sharp enough to describe the slope of loci inside $\mathcal{A}_g$, i.e. we know that any modular form with slope $&lt; q_1$ vanishes on a certain locus while any modular form with slope $> q_2$ does not vanish on the same locus, and there is no modular form with slope $q_1 &lt; q &lt; q_2$. I expect the modular forms to be <em>sloppy</em> (sorry for the pun!) for this purpose, but I think the main question is of some interest anyway.</p>

http://mathoverflow.net/questions/111173/how-long-can-a-primal-egyptian-fraction-be-that-optimally-approaches-unity/111220#111220 Stefano Pascolutti 2012-11-01T22:14:19Z 2012-11-01T22:14:19Z

I missed the part about <i>primary pseudoperfect numbers</i>. That will do the trick.

http://mathoverflow.net/questions/111173/how-long-can-a-primal-egyptian-fraction-be-that-optimally-approaches-unity/111220#111220 Stefano Pascolutti 2012-11-01T22:12:33Z 2012-11-01T22:12:33Z

I think this is the &quot;solution&quot; from above (see my comment). By the way, is you take the first eight factors of that number, you get a solution for the original problem for $n = 8$, which a step further.

http://mathoverflow.net/questions/111173/how-long-can-a-primal-egyptian-fraction-be-that-optimally-approaches-unity Stefano Pascolutti 2012-11-01T21:17:32Z 2012-11-01T21:17:32Z

Are you also interested in approximation from above? Would you accept 1/2 + 1/3 + 1/5 = 1 + 1/30? I don't have an answer for that, I am just curious.