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Old 2013-07-08, 23:46   #12
Citrix
 
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Looking more into this: Find bases such that base+1 and base-1 are not smooth. Otherwise you easily get trivial factors.


Related question: For each base find the lowest starting term that results in trivial factors. What is the lowest one for 2?
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Old 2013-07-09, 01:14   #13
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Code:
 
for(k=1,100000000000000,for(n=1,1000000000,if(isprime(k*2^n+1),print1(n",");k=k*2^n;break())))
is the brute force script I sent to Citrix through PM after this thread started, I know of a few tweaks I just can't get all them to work ( like eliminating n based on k, 2^n fall into two categories mod 6 2 and 4 so I tried to use ifs to control the n searched for each k but I can't get it to work). I know the multiplies by 2^n can be made into bitshift k<<n operations but that's about all I came with codes are interesting to look at.

Last fiddled with by science_man_88 on 2013-07-09 at 01:19
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Old 2013-07-09, 09:47   #14
henryzz
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With PFGW I don't really think it is worth bothering with that sort of thing. Almost all the time is spent trial factoring the candidates with large factors and prp testing them. Calculating the candidates is trivial.
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Old 2013-07-09, 14:21   #15
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Base 12 with prime b+1, b-1 ends in a trivial factor for k=1 c=+1
http://factordb.com/index.php?query=...at=1&sent=Show
Base 2 gets off lighter because it doesn't matter if every term is divisible by b-1=1.
b=12, k=1, c=-1 hasn't terminated yet for me.
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Old 2013-07-09, 15:36   #16
Citrix
 
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Quote:
Originally Posted by henryzz View Post
Base 12 with prime b+1, b-1 ends in a trivial factor for k=1 c=+1
http://factordb.com/index.php?query=...at=1&sent=Show
Base 2 gets off lighter because it doesn't matter if every term is divisible by b-1=1.
b=12, k=1, c=-1 hasn't terminated yet for me.
That is because of 11 being a factor of base-1.
FYI: I tested b=2, k=1, c=1 further upto 65,000 bits... no primes. This is how I did it with the (-f Flag)

Code:
ABC2 ((((((((((((((((1*2^1+1)*2^1+1)*2^2+1)*2^1+1)*2^5+1)*2^1+1)*2^1+1)*2^29+1)*2^3+1)*2^37+1)*2^31+1)*2^227+1)*2^835+1)*2^115+1)*2^7615+1)*2^6071+1)*2^$a+1
a: from 1 to 50000
If there was some way of sieving this series, we can go deeper.

Last fiddled with by Citrix on 2013-07-09 at 15:37
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Old 2013-07-31, 01:51   #17
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Quote:
Originally Posted by Citrix View Post
That is because of 11 being a factor of base-1.
FYI: I tested b=2, k=1, c=1 further upto 65,000 bits... no primes. This is how I did it with the (-f Flag)

Code:
ABC2 ((((((((((((((((1*2^1+1)*2^1+1)*2^2+1)*2^1+1)*2^5+1)*2^1+1)*2^1+1)*2^29+1)*2^3+1)*2^37+1)*2^31+1)*2^227+1)*2^835+1)*2^115+1)*2^7615+1)*2^6071+1)*2^$a+1
a: from 1 to 50000
If there was some way of sieving this series, we can go deeper.
FYI: I have finished this to n=130,000 with no primes....
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Old 2013-08-15, 22:15   #18
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Code:
((((((((((((((((1*2^1+1)*2^1+1)*2^2+1)*2^1+1)*2^5+1)*2^1+1)*2^1+1)*2^29+1)*2^3+1)*2^37+1)*2^31+1)*2^227+1)*2^835+1)*2^115+1)*2^7615+1)*2^6071+1)*2^218431+1 is prime!
Checked to n=40,000 after that.
The nash weight of this new number is even lower.

Last fiddled with by Citrix on 2013-08-15 at 22:16
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Old 2013-08-24, 05:24   #19
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Complete to 100K. continuing.
I have written a sieve for these numbers... so things are easier now.

Could one of the moderators move this thread to the open projects section. Thx
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