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Old 2019-04-19, 23:05   #188
R. Gerbicz
 
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"Robert Gerbicz"
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Quote:
Originally Posted by paulunderwood View Post
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.
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Old 2019-04-19, 23:19   #189
paulunderwood
 
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Quote:
Originally Posted by R. Gerbicz View Post
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
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Old 2019-04-19, 23:31   #190
Batalov
 
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"Serge"
Mar 2008
Phi(3,3^1118781+1)/3

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Quote:
Originally Posted by R. Gerbicz View Post
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.

I added a little friendly competition thread. Have some fun!
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Old 2019-04-20, 02:06   #191
rudy235
 
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Quote:
Originally Posted by paulunderwood View Post
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. 😋
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Old 2019-04-22, 02:47   #192
Batalov
 
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Phi(3,3^1118781+1)/3

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Quote:
Originally Posted by paulunderwood View Post
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...
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Old 2019-05-31, 10:21   #193
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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.
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Old 2019-09-11, 11:45   #194
paulunderwood
 
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Congrats to Ryan for a top20 prime 7*6^6772401+1 (5269954 digits)
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Old 2019-09-18, 16:59   #195
pepi37
 
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After milion nines:)

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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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