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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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Phi(4,2^7658614+1)/2

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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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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
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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
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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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Old 2020-03-04, 03:02   #196
rudy235
 
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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
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Old 2020-03-13, 15:21   #197
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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
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Old 2020-03-16, 17:13   #198
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Quote:
Originally Posted by paulunderwood View Post
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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