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2019-04-19, 23:05   #188
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

1,327 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

43×73 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

3·5·599 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/.

947 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

100011000110012 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/. 94710 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 313910 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:) 121610 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
 2020-03-04, 03:02 #196 rudy235     Jun 2015 Vallejo, CA/. 947 Posts This one has not been yet verified but it looks genuine. 6962 · 312863120 - 1 4269952 Digits. Largest of the year. Will rank 20 if verified. https://primes.utm.edu/primes/page.php?id=130702 Last fiddled with by VBCurtis on 2020-03-04 at 03:06 Reason: Fixed exponent rendering
 2020-03-13, 15:21 #197 paulunderwood     Sep 2002 Database er0rr 313910 Posts Congrats to Ryan Propper for his recent batch of proth mega primes for k = 9, 11 and 13 the largest of which has 3,462,100 digits Last fiddled with by paulunderwood on 2020-03-13 at 15:22
2020-03-16, 17:13   #198
JeppeSN

"Jeppe"
Jan 2016
Denmark

6516 Posts

Quote:
 Originally Posted by paulunderwood Congrats to Ryan Propper for his recent batch of proth mega primes for k = 9, 11 and 13 the largest of which has 3,462,100 digits
What ranges was Propper searching, and did he keep the residues of all the composite candidates he must have covered? This information could be useful to PrimeGrid which is planning to search and double-check (at least a part of) these Proth number regions, see https://www.primegrid.com/stats_div_llr.php.

/JeppeSN

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