... Yes, that's eminently possible: it is easy to make the cofactors with digits/4 into simple quintics or sextics. The local problem is that with my GGNFS...
Let m=67440294559676054016000 y=(m*(10^71+145589)+22)^2 N=(y-11^4)*(y-35^2)*(y-47^2)*(y-94^2)*(y-146^2)*(y-148^2)/m-5 then N+k is factorized for k in [0,5]. ...
... Congratulations! Very neat. I was wondering whether you would go for 6 record numbers with a construction inspired by Jaroslaw's P(X). Thanks for showing...
... The current status page shows a 10043-digit AP4 by Ken Davis: (97070894 + 104086947*n)*2^33333+1, n=0..3 It's the second largest known with unfortunate...
Hello ! I have a collection about 210 6k twins. for n+2 and n-2 I have scan all primfactors up to 1.6E9, without success. Perhaps find one of you the 4th...
... updated ... Indeed. Started just after Mike Oakes 10004 digit AP4 discovery in december http://tech.groups.yahoo.com/group/primeform/message/8092 I've...
... If you run ECM up to p20 level you have a good chance: mu = 2*210*exp(Euler)*20/6200 = 2.4 and then you may well end up with a long haul, usin ECPP at 6.2k...
... Around half would find no factor from 1.6E9 to p20 level ECM, so that half doesn't contribute to the chance. And p20 ECM on 420 6k numbers is not easy: ...
Hello ! That is the problem ! A ECM test for a single number ( I have 420!) takes to many time. Better is a Pollard Test for 6k numbers but I don't have a...
... I don't know why GMP-ECM stopped giving recommended p15 parameters. ... This is the README file of gmp-ecm-5.0, a new version of gmp-ecm, replacing version...
... Congratulations! http://hjem.get2net.dk/jka/math/consecutive_factorizations.htm is updated. The algebraic cofactors have been upgraded from "relatively...
Let m=67440294559676054016000 y=(m*(10^96+10624986)+22)^2 N=(y-11^4)*(y-35^2)*(y-47^2)*(y-94^2)*(y-146^2)*(y-148^2)/m-4 then N+k is factorized for k in [0,6]. ...
Congrats to Jens for http://primes.utm.edu/primes/page.php?id=82059 David PS: Chris will need to add Jens's comments to complete the recording of the AP3....
... Thanks. I had some time between other projects and got a lucky hit at 11% of the expectation. I don't know whether there is a submitter procedure for AP3...
... Congratulations! I've added the necessary comments on the other primes. (And you are right--manual intervention was required, but well worth it for a new...
... Yes, an agreeable swap! I was somewhat lucky: I had 12 chances of extending (or filling) 1400-digit constellations of 6 factorizations to get k=7. Running...
... Yes, congratulations. They found two cases with p12 and p13 as penultimate factors. There is no plan to prove prp6203. After 3 weeks with the page, my 4...
... I fixed it, at last. So now I guess that I am morally obliged to squeeze at least 25% extra digits. I'll try k=6 first, since there I have found a pair of ...
... That's easy to explain: your idea for a record page was so clearly and cleanly executed that anyone who had anything up their sleeve was almost bound to...
... Well, it may be that k=3 is also under threat: http://www.primegrid.com/all_news.php#11 ... But maybe that message is about a Woodall/Cullen find? David...
Breaking news: Raanan Chermoni discovered 2nd AP24: 1564588127269043 + 1249750*23#*n, for n = 0 to 23. Congratulations for this impressive discovery !!! ...
Hello group ! The new URL is http://anthony.d.forbes.googlepages.com/ktuplets.htm regards Norman Wissenswertes für Bastler und Hobby Handwerker. BE A BETTER...
