20120514, 10:02  #1  
"Frank <^>"
Dec 2004
CDP Janesville
4112_{8} Posts 
Escaping drivers (moved from 4788 thread)
Quote:
Quote:
I have two immediate examples that I recently dealt with vis a vis 2^9 * 3 * 11 * 31. First, 363270 broke this way: Driver acquired on line 1418, size 64, driver lost on line 1619, size 133; the lines after escape factored this way: Code:
2^9 * 3 * 11 * 31^2 2^8 * 3 * 11 * 31 2^8 * 3 2^8 * 3 2^8 * 3 2^8 * 3 2^8 * 3 2^8 * 3^4 * 31 2^8 * 3^3 * 31 2^8 * 3^3 2^9 * 3^2 2^7 * 3^2 2^7 * 3^4 2^7 * 3^2 2^7 * 3^2 2^7 * 3^2 2^7 * 3^2 2^8 * 3^2 2^8 * 3^5 2^11 * 3^5 2^11 * 3^6 2^10 * 3^5 2^10 * 3^5 2^7 For 572000, driver acquired on line 2886, size 86; driver lost on line 3070, size 151; after loss: Code:
2^9 * 3 * 11 * 31^2 2^8 * 3 * 11 * 31^2 2^8 * 3 * 11 * 31 2^8 * 3 * 11^3 * 31 2^8 * 3 * 11 2^8 * 3 2^8 * 3 2^8 * 3 2^8 * 3 2^6 * 3 2^6 * 3 2^6 * 3 2^6 * 3 2^4 * 3 2^4 * 3 2^4 * 3 2^4 * 3 2^4 * 3 2^4 * 3 2^4 * 3^2 2^4 * 7 

20120515, 18:15  #2 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2^{2}×3×5×97 Posts 
The lower the factor the more likely it is to stay on.
For 3 there is 50% chance of each factor keeping the 3 for the next iteration. The probability of keeping 3 with n factors is: Code:
n probability 1 50% 2 75% 3 87.5% 4 93.75% 5 96.88% Code:
n probability 1 16.67% 2 30.56% 3 42.13% 4 51.77% 5 59.81% I wonder if there is a formula estimating the number of factors a number of a certain size will have. 
20120515, 19:44  #3 
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
x^n for prime x either has 1 factor or n factors if n then x=2 gives you the most a given length can have, 2^3=8<10 so the maximum for 1 digits is 3 2^6=64<100 gives you up to 6 but it will change by from just 3 added each time because is about 3.321928094887362347870319430 this will round up another one every so often. your minimum the lowest prime below 10 is 7 I would believe that should give you a low end.

20120515, 20:53  #4  
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} Posts 
Quote:


20120518, 13:08  #5 
Romulan Interpreter
Jun 2011
Thailand
5^{2}·7·53 Posts 
Battlefield, episode 95280, challenging the record for breaking D4 ...
(related to the discussion in the 4788 thread) Code:
.1597 C137 = 2^4 * 3 * 5 * 31 * ... .1598 C137 = 2^4 * 3 * 5 * 31 * ... .1599 C138 = 2^4 * 3^3 * 5 * 31 * ... .1600 C138 = 2^4 * 3^3 * 5 * 31 * ... .1601 C139 = 2^4 * 3^2 * 5 * 7^2 * 31^3 * ... .1602 C139 = 2^4 * 3^2 * 5 * 7^2 * 31 * ... * C133 
20120519, 18:36  #6 
May 2009
Dedham Massachusetts USA
1101001011_{2} Posts 
Interestingly enough I was trying to figure this out. I went through the sequence from 276 to 5 million that reached 20 digits up to the merge or open end. I calculated the driver, number of digits and whether it stayed with the driver or escaped.
I think made a ratio of stayed/escaped up to 110 digits and did a linear fit of the line, a + bx where x is the number of digits. Note that this only gives a a very rough probability, because it weighs the less dependable high values the same as the lower points with many more datapoints. So stopping at different values could give the b for the down=driver from 1.4 to 1.3. This gives the following: 2^6*127: 127 + 4.6x 2*3: 3 + 4.93x 2^4*31: 31 + 3.06x 2^2*7: 7+ 3.38x 2^3*3*5: 17 + 1.21x 2^5*3*7: 8 + 1.05x downdriver: 2 + 1.36x 2^3*3: 2 + 0.39x 2^3*3^3*5 : 2 + 0.25 3^5*3^2*7: 1 + 0.27 For ones based on perfect numbers I chose the intecept values. Without this the values were close but not exact  basically the need for 2^6*127 to escape it requires 127^2 which only happens 1/127 even for 0 digits. These are in order of hardest to easiest to escape. Note that the 2^6 may not have had enough data to be very accurate. For the last driver 2^9*11*31, there were too many 0 exits to calculate this way. I separated the 2^3*3*5 and 2^5*3*7 by whether the power of 3 was 1 or greater than 1. The reason is that the power of 3 cannot be reduced easily because there are 2 powers of 3 from 2^3*5 (15,6) and from 2^5 (63) terms. 2*3*3^2*5 can only go to 2^3*3*5 if the power of 5 is higher than 1 (same as escaping) and 2^5*3^2*7 cannot go to 2^5*3*7 without changing the power of 2. My opinion (supported by the numbers above) is that 2^3*3^2*5 and 2^5*3^2*7 are not drivers. Interestingly, 2^3*3 had a much larger chance of escape than I expected compared to the downdriver. I guess the power of 3 can be higher than 1 adds more chance to escape. 
20120519, 18:46  #7 
May 2009
Dedham Massachusetts USA
3·281 Posts 
At 130 digits, 2^4*31 would have 1/430 chance of escape.
It occurs to me another definition of a driver could be based on whether the b from a+bx is greater than or equal 1, rather than the 2 deficit or surplus. In that case the drivers would be: 2^6*127 2*3 2^4*31 2^2*7 2^3*3^1*5 2^5*3^1*7 downdriver Higher perfect numbers (2^9*3*11*31 likely being a driver as well) but not 2*3*3, 2^3*3^2*5 or 2^5*3*7 which have 3x or greater times the chance to escape. 
20120519, 19:44  #8  
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} Posts 
Quote:
take x=25 as an example: 6p+31 if we want this next line to be divisible by 25 what properties of p are needed ? well 3125 = 6 so the number left over is 6*(p+1) so p+1 must be a multiple of 6 that is also divisible by 25 gcd(25,6) =1 so p+1 has a minimum of 25 leaving p = 24 mod 25 since half of these are even the p must be 49 mod 50 so for this to persist the prime must be of that form, failure of that form to contain primes shows if the driver can persist across lines. Last fiddled with by science_man_88 on 20120519 at 19:46 

20120520, 19:01  #9  
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 
Quote:
Code:
for(x=7,100,forprime(p=1,1000,print1(((sigma(x)x)*p+sigma(x))==(sigma(x*p)(x*p))==0));print(","x)) 

20120521, 21:58  #10  
"Forget I exist"
Jul 2009
Dumbassville
8384_{10} Posts 
Quote:


20120522, 10:50  #11 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2^{2}·3·5·97 Posts 
As far as I can work out the chance of losing the p in the next iteration after 2^x*p*comp is(assuming not a driver):
p=3 Code:
Digits 1 in x of losing 1 in x of keeping 60 18.46 1.057 70 19.93 1.053 80 21.30 1.049 90 22.59 1.046 100 23.81 1.044 110 24.97 1.042 120 26.07 1.040 130 27.14 1.038 140 28.16 1.036 160 30.10 1.034 180 31.92 1.032 Code:
Digits 1 in x of losing 1 in x of keeping 60 2.628 1.614 70 2.696 1.590 80 2.756 1.569 90 2.811 1.552 100 2.860 1.538 110 2.906 1.525 120 2.948 1.513 130 2.988 1.503 140 3.025 1.494 160 3.093 1.478 180 3.154 1.464 
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