20190419, 23:05  #188  
"Robert Gerbicz"
Oct 2005
Hungary
2×3^{2}×73 Posts 
Quote:
Say 2^a divides p1 3^b divides p+5 ofcourse you want 2^a~3^b (~sqrt(N)) because then you know large p1,(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 

20190419, 23:19  #189  
Sep 2002
Database er0rr
2^{2}·5^{2}·31 Posts 
Quote:
Last fiddled with by paulunderwood on 20190419 at 23:22 

20190419, 23:31  #190  
"Serge"
Mar 2008
Phi(3,3^1118781+1)/3
230A_{16} Posts 
Quote:
Surely now, 10 years later, one can repeat Ken's trick to find a couple 40,000digit sexy primes. I added a little friendly competition thread. Have some fun! 

20190420, 02:06  #191  
Jun 2015
Vallejo, CA/.
23×41 Posts 
Quote:
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. 😋 

20190422, 02:47  #192  
"Serge"
Mar 2008
Phi(3,3^1118781+1)/3
230A_{16} Posts 
Quote:


20190911, 11:45  #194 
Sep 2002
Database er0rr
2^{2}×5^{2}×31 Posts 
Congrats to Ryan for a top20 prime 7*6^6772401+1 (5269954 digits)

20190918, 16:59  #195 
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 20190918 at 16:59 
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