20200219, 19:16  #144 
Jul 2003
So Cal
807_{16} Posts 
As behind as the status page suggests. In the weeds.
This is why I've been prioritizing the base2 numbers. But 2,1165+ will give us some time to catch up a bit. And I don't think anyone is in a hurry to know these factors. They'll get done eventually. If anyone wants to solve a 70M+ matrix, send them my way! 
20200220, 00:30  #145 
"Curtis"
Feb 2005
Riverside, CA
4614_{10} Posts 
I'll do any matrix around 60M for your queue; if you stumble into one in the low 60s, give me a holla.

20200221, 14:42  #146 
Nov 2003
16444_{8} Posts 
A New Target (easy!)

20200221, 23:58  #147  
Jun 2012
3^{3}·109 Posts 
Quote:
Code:
2076486865904164187880498803002833020624706055858258295123907760787910463183237701437319913688727165276132151609318284002818920807675158414601157967453931895433506042829474274993772412901816590191592923 

20200222, 00:12  #148 
"Curtis"
Feb 2005
Riverside, CA
4614_{10} Posts 
Did it have a t60 worth of t65sized curves? I mean, is any more ECM necessary?
Sean and I can poly select this within a couple weeks. We could imitate the 2,1165+ sieve approach, using CADO for Q under, say, 100M and the 15e queue for 100Mup. Or just a Spring teamCADOsieve with A=30 (equivalent to I=15.5), which would need about 5GB ram per process. Last fiddled with by VBCurtis on 20200222 at 00:14 
20200222, 00:53  #149  
Nov 2003
2^{2}·5·373 Posts 
Quote:
my work plus the work of others [extent unknown], it has had more than sufficient ECM. The total extent is unknown: too many different participants, each with an unknown amount of work. I do believe that Bruce did a t65 by himself. It was among the first 5 holes when he did his work. 

20200222, 02:02  #150  
Jun 2012
3^{3}×109 Posts 
Quote:
Last fiddled with by swellman on 20200222 at 02:47 

20200222, 07:26  #151  
"Curtis"
Feb 2005
Riverside, CA
2·3·769 Posts 
Quote:
I think I'd pick 33/34LP if it were a pure CADO job, so going down half a largeprime to be compatible with the 15e queue is no big deal. Something like Q=5150M on CADO and 150600 on 15e ought to do the trick. 

20200222, 07:40  #152 
Jun 2003
2·3^{2}·269 Posts 

20200222, 08:47  #153 
"Carlos Pinho"
Oct 2011
Milton Keynes, UK
2^{4}×3×101 Posts 
On 2,1165+ I had a wonderful feedback from teams. They advise setting up a new app with details on memory requirements on project preference page and increase reward. I believe this is feasible, only maybe change or add more intermediate badge levels. Right now individuals cannot reach highest badge level.

20200222, 13:00  #154  
Nov 2003
2^{2}×5×373 Posts 
Quote:
The total number of lattice points that are sieved is minimized when the sieve area for each q is proportional to the yield for that q. The constant of proportionality falls out of the analysis as an eigenvalue in the calc of variations problem. Its value depends on the total number of relations needed. Since smaller q have higher yields this means that the sieve area for small q should be larger. One would think that smaller q would have smaller yield, but the following happens: There is a "seesaw" effect that takes place between the two norms that need to be smooth. As one makes one norm smaller (let's say the rational one), the other norm gets bigger [and vice versa]. The effect is nonlinear; the rising norm increases faster than the decreasing norm decreases. To see this look at what happens for a fixed factorization when one changes the algebraic degree. Also, look at what happens as q changes size. For example, we need (rational norm/q) to be smooth as q changes. As q gets bigger this gets smaller. But the algebraic norm increases as ~q^(d/2) where d is the degree when we use q on the rational side. 

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