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Old 2018-07-24, 06:46   #1
enzocreti
 
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Default PG PRIMES

https://solutionsti360.ca/MATH/pfgw/stats.html
At this link you can join in the ec-primes search.
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Old 2018-07-24, 08:35   #2
henryzz
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Quote:
Originally Posted by enzocreti View Post
https://solutionsti360.ca/MATH/pfgw/stats.html
At this link you can join in the ec-primes search.
What software is being used for the prp tests? It would be possible to use pfgw based on a script for this form.
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Old 2018-07-24, 08:45   #3
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Originally Posted by henryzz View Post
What software is being used for the prp tests? It would be possible to use pfgw based on a script for this form.
They are using pfgw. Do you know if it is multithreaded, can’t see anything on the txt documents.
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Old 2018-07-25, 06:12   #4
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Why is this project so slow?

Even one computer is enough to quickly find:
(2^75894-1)*10^22847+2^75893-1
(2^79798-1)*10^24022+2^79797-1
(2^92020-1)*10^27701+2^92019-1
...

What's more - these are known to FactorDB for months (since March).
________________

Ah, there it is https://oeis.org/A301806


2, 3, 4, 7, 8, 12, 19, 22, 36, 46, 51, 67, 79, 215, 359, 394, 451, 1323, 2131, 3336, 3371, 6231, 19179, 39699, 51456, 56238, 69660, 75894, 79798, 92020, 174968, 176006, 181015, 285019, 331259

Last fiddled with by Batalov on 2018-07-25 at 07:03
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Old 2018-07-25, 09:26   #5
enzocreti
 
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Quote:
Originally Posted by Batalov View Post
Why is this project so slow?

Even one computer is enough to quickly find:
(2^75894-1)*10^22847+2^75893-1
(2^79798-1)*10^24022+2^79797-1
(2^92020-1)*10^27701+2^92019-1
...

What's more - these are known to FactorDB for months (since March).
________________

Ah, there it is https://oeis.org/A301806


2, 3, 4, 7, 8, 12, 19, 22, 36, 46, 51, 67, 79, 215, 359, 394, 451, 1323, 2131, 3336, 3371, 6231, 19179, 39699, 51456, 56238, 69660, 75894, 79798, 92020, 174968, 176006, 181015, 285019, 331259

yes I found that values, but I have yet not found a probable prime congruent to 6 (mod 7)
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Old 2018-07-25, 15:01   #6
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So, these are not EC primes.

Rather they are PG primes, aren't they?
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Old 2018-07-26, 06:41   #7
enzocreti
 
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Quote:
Originally Posted by Batalov View Post
So, these are not EC primes.

Rather they are PG primes, aren't they?

Yes Enzo Creti is just a pseudonime
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Old 2019-05-07, 12:57   #8
enzocreti
 
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Default pg primes !!!

pg(k) are numbers of the form =(2^k-1)*10^d+2^(k-1)-1 where d is the number of decimal digits of 2^(k-1)-1.


The k for wich pg(k) is prime are:


2, 3, 4, 7, 8, 12, 19, 22, 36, 46, 51, 67, 79, 215, 359, 394, 451, 1323, 2131, 3336, 3371, 6231, 19179, 39699, 51456, 56238, 69660, 75894, 79798, 92020, 174968, 176006, 181015, 285019, 331259, 360787, 366770


I call a(1)=2, a(2)=3 a(3)=4 a(4)=7...the terms of the sequence


Now look at a(7s)'s


a(7)=19 is prime
a(14)=215 is a multiple of 215
a(21)=3371 is prime
a(28)=75894 is -1 (mod 215)
a(35)=331259 is prime


They seem not random at all!!!
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Old 2019-05-07, 14:10   #9
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I moved this to your blog area.

Keep your stuff there. You have been previously warned to not post your every thought elsewhere in the forum. If you don't behave, you can lose your posting privileges.

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Old 2019-05-07, 15:17   #10
enzocreti
 
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Quote:
Originally Posted by Uncwilly View Post
I moved this to your blog area.

Keep your stuff there. You have been previously warned to not post your every thought elsewhere in the forum. If you don't behave, you can lose your posting privileges.

sorry
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Old 2019-05-07, 15:32   #11
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Quote:
Originally Posted by enzocreti View Post
pg(k) are numbers of the form =(2^k-1)*10^d+2^(k-1)-1 where d is the number of decimal digits of 2^(k-1)-1.
Reducing the number of variables (but increasing the complexity), this is:

(2^k-1)(10^ {\lfloor{1+log_{10}{(2^{k-1}-1)}}\rfloor})+2^{k-1}-1
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