Quote:
Originally Posted by dave_dm
Has anyone looked into working out such an algorithm?

I tried to work this out, I am not an expert so there may be some point I have missed:
Say you have done the expected number of curves to bound B1=m (and saved the stage one residues), and the next higher digit size d has optimal bound B1=n.
Let p,q be the probability of one curve finding a factor of size d with B1=m,B1=n respectively. Let r,s,t be the time to complete stage one to B1=m, stage one to B1=n, and stage two from B1=n respectively.
Then it is worth continuing from the saved residue if (qp)/(sr+t) > q/(s+t), or equivalently if the ratio (qp)(s+t) / q(sr+t) is greater than 1.
As an example I tried curves on 2^9711 with a P4 2.66GHz and Prime95 (B2=100B1) using n=11e6, m=43e6. I got the probabilities p=1/79353, q=1/18461 by using Richard McIntosh's Maple program <http://www.mersenneforum.org/showpo...06&postcount=19> with factor size 50.5, and the resulting times were r=151, s=588, t=289 seconds. The ratio came to 0.927, so in this case it is better to start a fresh curve than continuing from the saved one.