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 Register FAQ Search Today's Posts Mark Forums Read  2015-11-14, 08:20 #23 Dubslow Basketry That Evening!   "Bunslow the Bold" Jun 2011 40 2015-11-29, 09:38   #24
Dubslow

"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88

3·29·83 Posts Quote:
 Originally Posted by Dubslow 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 : print(nt.factor(n))
7927 * 12569 * 93187

In : aq.twos_count(n)
Out: 6

In : 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 : 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 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 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 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 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 (2s count = 3(1) = 3), or the product of one prime of form 4n+1 and another of form (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 always holds, so the test will always work. But a quick foiling of results in , where . That quickly renders such an additional test useless.
Quote:
 Originally Posted by Happy5214  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, .
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   2015-11-29, 14:26 #25 Dubslow Basketry That Evening!   "Bunslow the Bold" Jun 2011 40 2015-11-29, 16:51 #26 Dubslow Basketry That Evening!   "Bunslow the Bold" Jun 2011 40 2015-11-30, 07:29   #27
henryzz
Just call me Henry

"David"
Sep 2007
Cambridge (GMT)

2×2,837 Posts Quote:
 Originally Posted by Dubslow 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   2015-11-30, 16:43   #28
Dubslow

"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88

3·29·83 Posts Quote:
 Originally Posted by henryzz @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.   2015-11-30, 17:05   #29
henryzz
Just call me Henry

"David"
Sep 2007
Cambridge (GMT)

2·2,837 Posts Quote:
 Originally Posted by Dubslow 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   2015-11-30, 17:37   #30
Dubslow

"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88

160658 Posts Quote:
 Originally Posted by henryzz 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   2015-11-30, 22:56 #31 Dubslow Basketry That Evening!   "Bunslow the Bold" Jun 2011 40 2015-12-02, 19:30   #32
henryzz
Just call me Henry

"David"
Sep 2007
Cambridge (GMT)

2·2,837 Posts Quote:
 Originally Posted by henryzz 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   2016-10-20, 14:40 #33 Dubslow Basketry That Evening!   "Bunslow the Bold" Jun 2011 40 Thread Tools Show Printable Version Email this Page

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