20150402, 17:27  #243 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
23DF_{16} Posts 
D.Broadhurst undoubtedly has tabulated all of those that hold promise of a KP or a CHG proof* as long ago as a decade ago, so all easy ones are done! The usual contributors in this particular class are Chuck Lasher, Bouk de Water and David himself. PRPtop has some PRP leads and I've recently added some by PRPscanning primU/V()'s up to n<400,000. Note: for prime N, primU() = simply Fib() and primV() = simply lucas(), so these were known for years (and found by Lifchitz).
Note: D.Broadhurst also maintains the ad hoc reservation for the "algebraically hopeless" primU/V()'s, so if you would be thinking about running Primo on one, you will do well by first reserving it. (I didn't know that, but now I do, and I spread the word.) _______________ *That promise is in factoring at least some of the composites in the N1 or N+1 factorization. See primU(137439)1 cofactors; they can yield to ECM or in some cases NFS; also note that for some of them mersennus.net/fibonacci may have a better factorization that was at some time imported into FactorDB. Last fiddled with by Batalov on 20150402 at 17:34 Reason: typos, footnote 
20150403, 15:02  #244 
Sep 2009
1948_{10} Posts 
Continuing my search for PRPs with factorable N+1 or N1 I've found http://factorization.ath.cx/index.ph...00000635654313, (10^1000+25739)^10*10+1, where N1 has a large PRP factor http://factorization.ath.cx/index.ph...00000746266343 to the 10th power. So proving it will make N1 fully factored.
Chris 
20150404, 15:40  #245 
Sep 2009
2^{2}×487 Posts 
@Batalov, thanks for that, so I didn't miss anything significant. So need not waste time looking for something that's not there.
Chris 
20150405, 19:59  #246 
Sep 2009
79C_{16} Posts 
And another like the last one:
Proving http://factorization.ath.cx/index.ph...00000772489376 (1007 digits) will enable a N1 proof for http://factorization.ath.cx/index.ph...00000635678150, (10^1015+38358)^10*10+1 (10152 digits). Chris 
20150407, 17:09  #247 
Sep 2009
11110011100_{2} Posts 
So near and yet so far.
After adding algebraic factors (2^34861*391)/771 (10494digits) is now 22.099% factored, and it also has one large PRP http://factorization.ath.cx/index.ph...00000772818863 (1180 digits). Which is not quite enough for factordb to prove (2^34861*391)/77 prime.
And no other composite factors are small enough to be worth trying to factor. It's 2^348601 after removing small factors so probably has had enough ECM etc run against it's factors that finding another factor would take more work that proving (2^34861*391)/77 prime with PRIMO. Chris 
20150407, 19:04  #248  
Nov 2012
2^{3}·3^{2} Posts 
Quote:


20150407, 19:33  #249  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
10AB_{16} Posts 
Quote:
Code:
from math import log10 whole_number = log10((2**34861*391)//771) factored_ratio = whole_number * 0.22099 additional_prime = log10(3506558761926971512477297296683018240237802676731644055250045204537775194400010178087998447321507478600817592705616161405817738352734894144721317929263324912197266117511551580761732541869669876689987165726621178684600171482007104597962677886695675167576338925575553986046610303368144884568916407098842880063824741179033940355870192625557343740123589410849730577079283148102933868035818491353137898013930244384291688588275054632147751736981106084248913391524320601903576237544952769251891260207999716952903190778749065796352828097883738968079636783614385584428272648346785025793447800638665413535803201979495988829833362404777758623082617387265333067529872148141192233553075004206944147802358620404080736497689698476221395097842352160492710593555790826888291012451038050049824479137168165964693849277060186810651364706931928084605931133410403663856663152305108447186454974035882713078010346173759344206193326603356739063396771113048615128471201410957739917292042987815135728382012015386609272490787328671741261167085118548057393212461278670329195501025353396663783059566063839318304945626306880713050813110162727173939454863315799108485675073025126029693463381547229527938485439361) # or additional_prime = 1180 + log10(.350655876) # the length plus log10([dot]the starting digits of the long number) new_factored_ratio = additional_prime + factored_ratio print(new_factored_ratio/whole_number) Last fiddled with by MiniGeek on 20150407 at 19:41 

20150408, 16:00  #250 
Sep 2009
3634_{8} Posts 
OK, Thanks for proving it. I'll make a note that the limit's exactly 33.3333% factored.
So using bc I could have done: 1180*100/10494+22.099 33.343 But there are a lot of numbers with N+1 or N1 between 25% and 33.333% factored. @syd, could you please upgrade the test to work in this case? Chris 
20150408, 18:03  #251 
"William"
May 2003
New Haven
2^{3}×5×59 Posts 

20150408, 18:05  #252 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
21737_{8} Posts 
Running KP tests (i.e. for 30% ... 1/3 factored N^21) would be easy to set up; this is just a fairly simple sequence of scripts, but ~28% ...30% will have difficulty running unattended, and some of them may be quite long. Some of the tougher test (~26% ...28% CHG) may run for a very, very, very long time. Not a robust idea on a server.
There is a straightforward KP implementation as a GP script on top of the usual PFGW test (which will run and report PRP with 90%+ proof). 
20150408, 18:13  #253 
Sep 2002
Database er0rr
5×701 Posts 

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