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Old 2015-11-14, 08:20   #23
Dubslow
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Code:
bill@Gravemind⌚0211 ~/bin/py/numtheory ∰∂ scripts/drivers.py 
Getting the current data
Sequence 383320 has a driver but also has class 3: 2^3 * 3^2 * 5^2 * C121
Sequence 661998 has a driver but also has class 3: 2^3 * 3^2 * 5^2 * C122
Sequence 290160 has a driver but also has class 3: 2^3 * 3^2 * 5^2 * C124
Sequence 258450 has a driver but also has class 3: 2^3 * 3^2 * 5^2 * C128
Sequence  81192 has a driver but also has class 3: 2^5 * 3 * 7^2 * C122
Sequence 151032 has a driver but also has class 3: 2^5 * 3 * 7^2 * C120
Sequence  48780 has a driver but also has class 3: 2^3 * 3^2 * 5^2 * C140
240810 may have a driver that's ready to break (composite is 1 mod 4): 2^3 * 3^2 * 5 * C122
976218 may have a driver that's ready to break (composite is 1 mod 4): 2^3 * 3^4 * 5 * C123
971496 may have a driver that's ready to break (composite is 1 mod 4): 2^3 * 3^2 * 5 * C124
 50892 may have a driver that's ready to break (composite is 1 mod 4): 2^2 * 7^2 * C125
360876 may have a driver that's ready to break (composite is 1 mod 4): 2^2 * 7^2 * C126
845844 may have a driver that's ready to break (composite is 1 mod 4): 2^3 * 3^4 * 5 * C126
223398 may have a driver that's ready to break (composite is 1 mod 4): 2^2 * 7^2 * C126
 48510 may have a driver that's ready to break (composite is 1 mod 4): 2^3 * 3^2 * 5 * C126
838320 may have a driver that's ready to break (composite is 1 mod 4): 2^3 * 3^2 * 5 * C127
104286 may have a driver that's ready to break (composite is 1 mod 4): 2^3 * 3^2 * 5 * C127
943182 may have a driver that's ready to break (composite is 1 mod 4): 2^3 * 3^2 * 5 * C127
752970 may have a driver that's ready to break (composite is 1 mod 4): 2^3 * 3^2 * 5 * C133
999558 may have a driver that's ready to break (composite is 1 mod 4): 2^2 * 7^2 * C137
880512 may have a driver that's ready to break (composite is 1 mod 4): 2^3 * 3^2 * 5 * C144
https://github.com/dubslow/MersenneF...pts/drivers.py
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Old 2015-11-29, 09:38   #24
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Quote:
Originally Posted by Dubslow View Post
aid with describing the cases to mutate a class 3 or 4 (or more) guide is appreciated.
Given my newfound understanding of what CStern's work actually means, I've written fully general mutation analysis code (requiring only the number to be tested, target twos count, and the form of the factors to be assumed):

Code:
In [32]: print(nt.factor(n))
7927 * 12569 * 93187

In [33]: aq.twos_count(n)
Out[33]: 6

In [34]: res=aq.possible_mutation(n, 6, [1,1,1]);  print(aq.possible_mutation_to_str(res,str(n)))
Assuming that n=9284636703581 is made of 3 primes, then since it's 1 (mod 4), it's possible that tau(n)=3=1+1+1 via the following conditions:
     p0%4==1, p1%4==1, p2%4==1
Assuming that n=9284636703581 is made of 3 primes, then since it's 5 (mod 8), it's possible that tau(n)=5=1+2+2 via the following conditions:
     p0%8==3, p1%8==3, p2%8==5
Assuming that n=9284636703581 is made of 3 primes, then since it's 13 (mod 16), it's possible that tau(n)=6=1+2+3 via the following conditions:
     p0%16==1, p1%16==7, p2%16==11
     p0%16==3, p1%16==7, p2%16==9

In [35]: for p in nt.factor(n):
    print('{} % {} = {}'.format(p, 16, p%16))
   ....:     
12569 % 16 = 9
93187 % 16 = 3
7927 % 16 = 7
Quote:
Originally Posted by axn View Post
2^3*3^2*5*p*q where both p and q are 1 (mod 4) will go from 2^3 to a *higher* power of 2. Because 5,p,and q will each contribute a 2 in terms of (5+1), (p+1) and (q+1), and hence the difference will be 2^3*odd - 2^3*odd = 2^3 * even.
Quote:
Originally Posted by Happy5214 View Post
I don't consider myself an expert in math, but I can use Wolfram|Alpha. My understanding, which may be incorrect (please correct me if it is), is that the 2s count of a composite number is the sum of the 2s counts of its factors. Also, the 2s count of a prime with a given power b is 0 if b is even, or s + c - 1 if b is odd, with s being the 2s count of the prime itself and c equaling the 2s count of the exponent. The 2s count of a prime p is the value s in the form s(2n + 1) of p + 1. Given that, a given cofactor and associated factors will break a driver when the 2s counts of the factors add up to a value less than or equal to the class of the driver.

Therefore, as we know, a class 2 driver will break if v is a prime of the form 4n+1 (2s count = 1), a prime of the form 8n+3 (2s count = 2), or a product of two primes, each of the form 4n+1 (2s count = 1+1 = 2). If I'm understanding this process right, a class 3 driver will break under any of the above circumstances, or if: v is a prime of the form 16n+7 (2s count = 3), the product of three primes of form 4n+1 (2s count = 3(1) = 3), or the product of one prime of form 4n+1 and another of form 8n+3 (2s count = 2+1 = 3). The break that occurs with a driver of class n when v is the product of ≤ n primes of form 4n+1 always holds, so the t \equiv 1 \pmod 4 test will always work. But a quick foiling of (4n+1)(8p+3) results in 4q+3, where q = 8np + 3n + 2p. That quickly renders such an additional test useless.
Quote:
Originally Posted by Happy5214 View Post


Obviously, I didn't mean to imply that c is the 2s count of the exponent when said exponent isn't prime itself. I should have said c is equal to n when representing the exponent in its Riesel form, k \cdot 2^n - 1.
I now understand these!

Happy, your analysis is indeed correct. Though I am curious, how did you come up with the twos count of prime powers? The proof isn't exactly trivial.

At any rate, I'll perform this analysis on all current sequences and post the results shortly.

Last fiddled with by Dubslow on 2015-11-29 at 09:39
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Old 2015-11-29, 14:26   #25
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These are the drivers that are prone to breaking. If you include non-drivers, then even excluding those with v=1, there are substantially more sequences which could mutate. And that's only assuming C is a semiprime. I expect some more sequences may be mutatable if C is made of three primes instead of two, but I didn't test that case.

Code:
240810 with guide 2^3 * 3^2 * 5 (class 2) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
976218 with guide 2^3 * 3^4 * 5 (class 2) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
971496 with guide 2^3 * 3^2 * 5 (class 2) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
 50892 with guide 2^2 * 7^2 (class 2) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
360876 with guide 2^2 * 7^2 (class 2) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
845844 with guide 2^3 * 3^4 * 5 (class 2) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
223398 with guide 2^2 * 7^2 (class 2) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
 48510 with guide 2^3 * 3^2 * 5 (class 2) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
943182 with guide 2^3 * 3^2 * 5 (class 2) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
838320 with guide 2^3 * 3^2 * 5 (class 2) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
104286 with guide 2^3 * 3^2 * 5 (class 2) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
752970 with guide 2^3 * 3^2 * 5 (class 2) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
999558 with guide 2^2 * 7^2 (class 2) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
880512 with guide 2^3 * 3^2 * 5 (class 2) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
 78474 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
835264 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
541458 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
589380 with guide 2^3 * 3^4 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
196848 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
224760 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
199152 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
329310 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
293040 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
151032 with guide 2^5 * 3 * 7^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
662016 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
504756 with guide 2^3 * 3^2 * 5^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
 31446 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
383320 with guide 2^3 * 3^2 * 5^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 3 (mod 8), it's possible that tau(n)=3=1+2 via the following conditions: p1%8==1, p2%8==3
756840 with guide 2^3 * 3^2 * 5^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
298464 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
595680 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 3 (mod 8), it's possible that tau(n)=3=1+2 via the following conditions: p1%8==1, p2%8==3
175968 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
310440 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 7 (mod 8), it's possible that tau(n)=3=1+2 via the following conditions: p1%8==3, p2%8==5
665064 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 3 (mod 8), it's possible that tau(n)=3=1+2 via the following conditions: p1%8==1, p2%8==3
511182 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
790308 with guide 2^5 * 3 * 7^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
584496 with guide 2^3 * 3^4 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
661998 with guide 2^3 * 3^2 * 5^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
 45570 with guide 2^3 * 3^4 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
 72408 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
977592 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 3 (mod 8), it's possible that tau(n)=3=1+2 via the following conditions: p1%8==1, p2%8==3
 81192 with guide 2^5 * 3 * 7^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
576150 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
214120 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 7 (mod 8), it's possible that tau(n)=3=1+2 via the following conditions: p1%8==3, p2%8==5
531024 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
141720 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
172554 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 3 (mod 8), it's possible that tau(n)=3=1+2 via the following conditions: p1%8==1, p2%8==3
290160 with guide 2^3 * 3^2 * 5^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
504810 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 3 (mod 8), it's possible that tau(n)=3=1+2 via the following conditions: p1%8==1, p2%8==3
129696 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
125034 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
897204 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
775230 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 3 (mod 8), it's possible that tau(n)=3=1+2 via the following conditions: p1%8==1, p2%8==3
671560 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
200022 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
229516 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
821232 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
788910 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
 69828 with guide 2^3 * 3^4 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 7 (mod 8), it's possible that tau(n)=3=1+2 via the following conditions: p1%8==3, p2%8==5
258450 with guide 2^3 * 3^2 * 5^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 3 (mod 8), it's possible that tau(n)=3=1+2 via the following conditions: p1%8==1, p2%8==3
111624 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
265776 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 7 (mod 8), it's possible that tau(n)=3=1+2 via the following conditions: p1%8==3, p2%8==5
293826 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
146964 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
 48780 with guide 2^3 * 3^2 * 5^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
966180 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 1 (mod 4), it's possible that tau(n)=2=1+1 via the following conditions: p1%4==1, p2%4==1
755250 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 3 (mod 8), it's possible that tau(n)=3=1+2 via the following conditions: p1%8==1, p2%8==3
  1464 with guide 2^3 * 3^2 (class 3) may mutate: Assuming that n=C is made of 2 primes, then since it's 3 (mod 8), it's possible that tau(n)=3=1+2 via the following conditions: p1%8==1, p2%8==3
434760 with guide 2^5 * 3^2 * 7^2 (class 5) may mutate: Assuming that n=C is made of 2 primes, then since it's 7 (mod 8), it's possible that tau(n)=3=1+2 via the following conditions: p1%8==3, p2%8==5

Last fiddled with by Dubslow on 2015-11-29 at 14:27
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Old 2015-11-29, 16:51   #26
Dubslow
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Here's an idea for a new subproject: the "standard" D5 is $2^5\cdot3\cdot7$ with class 0, but in fact $\sigma(2^6-1) = 3^2 \cdot 7$, so if a sequence mutates to $2^5 \cdot 3^2 \cdot 7$, then the $3^2$ is guaranteed to stay, so that the guide has class 2. We could target these for the subproject.

Last fiddled with by Dubslow on 2015-11-29 at 16:52
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Old 2015-11-30, 07:29   #27
henryzz
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Quote:
Originally Posted by Dubslow View Post
Here's an idea for a new subproject: the "standard" D5 is $2^5\cdot3\cdot7$ with class 0, but in fact $\sigma(2^6-1) = 3^2 \cdot 7$, so if a sequence mutates to $2^5 \cdot 3^2 \cdot 7$, then the $3^2$ is guaranteed to stay, so that the guide has class 2. We could target these for the subproject.
As far as I can tell there are 7 of them.
reserving one of them 696582
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Old 2015-11-30, 16:43   #28
Dubslow
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Quote:
Originally Posted by henryzz View Post
@Dubslow At first glance your script does three primes. Does this require a fair bit of computing power to run or something?
It might be nice to clear that list down.
I don't understand what you mean.
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Old 2015-11-30, 17:05   #29
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Quote:
Originally Posted by Dubslow View Post
I don't understand what you mean.
You only ran it for two primes.
I was basically asking why you didn't run it for three primes as your script appears to support 3 primes.

edit: I am referring to #27

Last fiddled with by henryzz on 2015-11-30 at 17:06
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Old 2015-11-30, 17:37   #30
Dubslow
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"Bunslow the Bold"
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40<A<43 -89<O<-88

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Quote:
Originally Posted by henryzz View Post
You only ran it for two primes.
I was basically asking why you didn't run it for three primes as your script appears to support 3 primes.

edit: I am referring to #27
Ah, I think you were missing an "n't do" in your original question. I probably should check for three primes as well... "soon".

Last fiddled with by Dubslow on 2015-11-30 at 18:10
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Old 2015-11-30, 22:56   #31
Dubslow
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"Bunslow the Bold"
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40<A<43 -89<O<-88

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There are no drivers that can be broken with 3 primes that couldn't also be broken with 2 primes. Nevertheless, perhaps they are more likely to break.


Last fiddled with by Dubslow on 2015-11-30 at 23:20 Reason: Thanks for the reminder sm88
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Old 2015-12-02, 19:30   #32
henryzz
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Quote:
Originally Posted by henryzz View Post
As far as I can tell there are 7 of them.
reserving one of them 696582
Strike 1
After rising 8 digits it is now on 2^7*3^2*5.
It is now stuck on a c128 with a full t40
releasing 696582

Looking at it now I wonder where I got the number 7 from there are many more.

reserving 888144
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Old 2016-10-20, 14:40   #33
Dubslow
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