20151114, 08:20  #23 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3×29×83 Posts 
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 
20151129, 09:38  #24  
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3·29·83 Posts 
Quote:
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:
Quote:
Quote:
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 20151129 at 09:39 

20151129, 14:26  #25 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
16065_{8} Posts 
These are the drivers that are prone to breaking. If you include nondrivers, 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 20151129 at 14:27 
20151129, 16:51  #26 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3×29×83 Posts 
Here's an idea for a new subproject: the "standard" D5 is $2^5\cdot3\cdot7$ with class 0, but in fact $\sigma(2^61) = 3^2 \cdot 7$, 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 20151129 at 16:52 
20151130, 07:29  #27  
Just call me Henry
"David"
Sep 2007
Cambridge (GMT)
2×2,837 Posts 
Quote:
reserving one of them 696582 

20151130, 16:43  #28 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3·29·83 Posts 

20151130, 17:05  #29 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT)
2·2,837 Posts 
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 20151130 at 17:06 
20151130, 17:37  #30 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
16065_{8} Posts 
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 20151130 at 18:10 
20151130, 22:56  #31 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
1C35_{16} Posts 
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 20151130 at 23:20 Reason: Thanks for the reminder sm88 
20151202, 19:30  #32  
Just call me Henry
"David"
Sep 2007
Cambridge (GMT)
2·2,837 Posts 
Quote:
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 

20161020, 14:40  #33 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3·29·83 Posts 
