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2019-04-19, 23:05   #188
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

2×32×73 Posts

Quote:
 Originally Posted by paulunderwood Congrats to GEN-ERIC for the primes [p,p+6] = (18041#/14*2^39003-4)±3. http://primepairs.com/
That is just painful. I mean the ecpp proof for the hard number when you have a much cheaper way, searching a better form.

Say
2^a divides p-1
3^b divides p+5

ofcourse you want 2^a~3^b (~sqrt(N)) because then you know large p-1,(p+6)-1 factors in the oder of sqrt(N), so

a=floor(log(N)/log(2)/2)
b=floor(log(N)/log(3)/2)

you can search p in the form (because 2^a and 3^b are coprime):

p=u*2^a+v*3^b,
from divisibilities you can get:

Code:
v*3^b==1 mod 2^a
u*2^a==-5 mod 3^b
Don't need to run two variables, fix u, then run v in arithmetic progression. Use sieving.

2019-04-19, 23:19   #189
paulunderwood

Sep 2002
Database er0rr

22·52·31 Posts

Quote:
 Originally Posted by R. Gerbicz That is just painful. I mean the ecpp proof for the hard number when you have a much cheaper way, searching a better form. Say 2^a divides p-1 3^b divides p+5 ofcourse you want 2^a~3^b (~sqrt(N)) because then you know large p-1,(p+6)-1 factors in the oder of sqrt(N), so a=floor(log(N)/log(2)/2) b=floor(log(N)/log(3)/2) you can search p in the form (because 2^a and 3^b are coprime): p=u*2^a+v*3^b, from divisibilities you can get: Code: v*3^b==1 mod 2^a u*2^a==-5 mod 3^b Don't need to run two variables, fix u, then run v in arithmetic progression. Use sieving.
Somebody, maybe GEN-ERIC with his 16 core Threadripper, should try to beat his record with the above method: The gauntlet has been thrown down!

Last fiddled with by paulunderwood on 2019-04-19 at 23:22

2019-04-19, 23:31   #190
Batalov

"Serge"
Mar 2008
Phi(3,3^1118781+1)/3

230A16 Posts

Quote:
 Originally Posted by R. Gerbicz That is just painful. I mean the ecpp proof for the hard number when you have a much cheaper way, searching a better form. Don't need to run two variables, fix u, then run v in arithmetic progression. Use sieving.
Exactly! This new thingy completely misses the precious beauty of Ken Davis' construction.
Surely now, 10 years later, one can repeat Ken's trick to find a couple 40,000-digit sexy primes.

2019-04-20, 02:06   #191
rudy235

Jun 2015
Vallejo, CA/.

23×41 Posts

Quote:
 Originally Posted by paulunderwood I'd argue Peter's triplet is not so sexy since there is a prime at 6521953289619 * 2^55555 - 1 On the other hand: there is a prime between 17 and 23

YepI And that is what we call a triplet, which -at least for large enough numbers- is more important than just a pair of sexy primes. 😋

2019-04-22, 02:47   #192
Batalov

"Serge"
Mar 2008
Phi(3,3^1118781+1)/3

230A16 Posts

Quote:
 Originally Posted by paulunderwood Congrats to GEN-ERIC for the primes [p,p+6] = (18041#/14*2^39003-4)±3. http://primepairs.com/
Well, that world record didn't live long...

 2019-05-31, 10:21 #193 rudy235     Jun 2015 Vallejo, CA/. 23×41 Posts A new Cunningham Chain of the 2nd kind was published a few days ago. Congratulations to Serge Batalov on the record. (2p+1) 556336461 · 2211356 - 1 with 63634 Digits HERE The previous record had 52726 digits.
 2019-09-11, 11:45 #194 paulunderwood     Sep 2002 Database er0rr 22×52×31 Posts Congrats to Ryan for a top20 prime 7*6^6772401+1 (5269954 digits)
 2019-09-18, 16:59 #195 pepi37     Dec 2011 After milion nines:) 1,201 Posts Small but sweet :) [Worker #1 Sep 18 17:41:47] 9*10^380734+1 is a probable prime! Wh8: 0976AF3D,00000000 And of course it is proven prime with LLR :) Last fiddled with by pepi37 on 2019-09-18 at 16:59

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