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 2013-02-15, 19:58 #1 ET_ Banned     "Luigi" Aug 2002 Team Italia 12D916 Posts Proth primes While working on GFNs (N=6000-6150, k=50,000,000-2,500,00,000), Markus Tervoonen gathered a huge list of Proth primes (about 40 millions). If you are interested please leave a message... Luigi
 2019-05-18, 05:48 #2 ATH Einyen     Dec 2003 Denmark 3,163 Posts Starting Proth prime test of 123173*2^333333+1 Using all-complex FMA3 FFT length 30K, Pass1=640, Pass2=48, clm=2, a = 3 123173*2^333333+1 is prime! (100349 decimal digits) Time : 46.491 sec. Starting Proth prime test of 182931*2^333333+1 Using all-complex FMA3 FFT length 30K, Pass1=640, Pass2=48, clm=2, a = 5 182931*2^333333+1 is prime! (100349 decimal digits) Time : 46.459 sec. Starting Proth prime test of 1460231*2^333333+1 Using zero-padded FMA3 FFT length 32K, Pass1=512, Pass2=64, clm=2, a = 3 1460231*2^333333+1 is prime! (100350 decimal digits) Time : 49.041 sec. Starting Proth prime test of 1569345*2^333333+1 Using zero-padded FMA3 FFT length 32K, Pass1=512, Pass2=64, clm=2, a = 11 1569345*2^333333+1 is prime! (100350 decimal digits) Time : 49.245 sec. Starting Proth prime test of 1714923*2^333333+1 Using zero-padded FMA3 FFT length 32K, Pass1=512, Pass2=64, clm=2, a = 5 1714923*2^333333+1 is prime! (100350 decimal digits) Time : 48.322 sec. Starting Proth prime test of 1751013*2^333333+1 Using zero-padded FMA3 FFT length 32K, Pass1=512, Pass2=64, clm=2, a = 5 1751013*2^333333+1 is prime! (100350 decimal digits) Time : 49.114 sec. Starting Proth prime test of 1852761*2^333333+1 Using zero-padded FMA3 FFT length 32K, Pass1=512, Pass2=64, clm=2, a = 5 1852761*2^333333+1 is prime! (100350 decimal digits) Time : 49.546 sec. Last fiddled with by ATH on 2019-05-19 at 16:49
 2019-07-31, 06:05 #3 Chara34122   Nov 2018 Russia 7 Posts I'm interested. Besides, today I've found one: 305147*2^1030527+1 (310226 digits long) from http://irvinemclean.com/maths/sierpin3.htm . How can I check it for being a Fermat, GF, xGF divisor?
2019-07-31, 12:23   #4
Dylan14

"Dylan"
Mar 2017

32·5·13 Posts

Quote:
 Originally Posted by Chara34122 I'm interested. Besides, today I've found one: 305147*2^1030527+1 (310226 digits long) from http://irvinemclean.com/maths/sierpin3.htm . How can I check it for being a Fermat, GF, xGF divisor?
You could use pfgw to test the number for Fermat divisibility. You’ll want to use the -gxo flag (which does the tests without testing to see if the number is prime). An appropriate command line for Windows would be

Code:
pfgw64 -gxo -q”305147*2^1030527+1”

 2019-08-28, 17:47 #5 Chara34122   Nov 2018 Russia 716 Posts If anyone is interested : one more 285473*2^530921+1 is prime! (159829 decimal digits).
 2019-09-09, 23:17 #6 rudy235     Jun 2015 Vallejo, CA/. 3·5·67 Posts A new prime has been discovered (pending verification) and has 5,269,954 digits. It is 7 ·6 6772401 + 1 . Although I am doubtful it was checked as a Proth Prime, it might be represented as such [7*3**6772401* 2**6772401 +1 . https://primes.utm.edu/primes/page.php?id=129914 It is by far the largest Prime of 2019 and if verified it will rank as #18 in the list of Largest Primes kept by CC Congratulations to Ryan Propper. Edit: A Proth number is restricted to k k>2n N=k2n+1 so I ammend my previous statement. Last fiddled with by rudy235 on 2019-09-10 at 00:15
 2019-09-11, 09:23 #7 pepi37     Dec 2011 After milion nines:) 23×3×61 Posts Congratulations to Ryan Propper. He really earn this prime: but I cannot even imagine what resources he has. If we make initial sieve , then assume sieve depth, from last prime we have at least 100000 candidates from 2.8M digits and above :) And he process 100000 those candidates in 34 days. Some supercomputer must be behind scene.
 2019-09-17, 16:45 #8 ixfd64 Bemusing Prompter     "Danny" Dec 2002 California 2×3×401 Posts He mentioned a few years ago that he had access to a cluster: https://mersenneforum.org/showthread.php?t=17690 I imagine he still does.
2020-02-15, 16:06   #9
storm5510
Random Account

Aug 2009

1,973 Posts

Quote:
 Originally Posted by rudy235 ...A Proth number is restricted to k k>2n N=k2n+1 so I ammend my previous statement.
Would the simple form of this not be k*2^n+1?

 2020-10-02, 07:11 #10 paulunderwood     Sep 2002 Database er0rr 380110 Posts Congrats to Ryan for the 17th largest known prime 7*2^18233956 + 1 with 5,488,969 decimal digits,

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