20091107, 17:04  #1 
Just call me Henry
"David"
Sep 2007
Liverpool (GMT/BST)
37×163 Posts 
Losing downdriver
I have been doing a little research into what driver the downdriver mutates into.
I started with 2*p. 50% If p = 3 mod 4 then we keep the downdriver. 50% If p = 1 mod 4 then the power of 2 is 2 or more: 25% If p = 1 mod 8 then the power of 2 is 2. It continues like this infinitely.12.5% If p = 5 mod 16 then the power of 2 is 3. I realize that most people know this but i like to see it put really simply. Expect 2*p*q soon(ish). 10metreh: That never loses the downdriver... Henryzz: Yes, My brain really wasn't switched on when i wrote this post. Last fiddled with by henryzz on 20091108 at 14:41 Reason: adding a little comment 
20091107, 17:18  #2  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
11×389 Posts 
Quote:
50% If p = 1 mod 4 then we still keep the downdriver. 50% If p = 3 mod 4 then the power of 2 is 2. And, more importantly: There are no primes p such that p = 6 mod 8, or any other similar form (i.e. p = x mod y with x and y even). It's quite obvious why: Any such p would have to be even, but since the only even prime is 2, and since 2 is not equal to 6 mod 8, no such primes exist. Last fiddled with by TimSorbet on 20091107 at 17:19 

20091107, 18:17  #3  
Just call me Henry
"David"
Sep 2007
Liverpool (GMT/BST)
37·163 Posts 
Quote:
i will correct it is a bit when i have time 

20091107, 18:40  #4 
Nov 2008
2×3^{3}×43 Posts 
Also, 1 mod 4 is the one that loses the downdriver, not 3 mod 4.
This should be better: 50% If p = 3 mod 4 then we keep the downdriver. 50% If p = 1 mod 4 then the power of 2 is 2 or more: 25% If p = 1 mod 8 then the power of 2 is 2. 12.5% If p = 5 mod 16 then the power of 2 is 3. Last fiddled with by 10metreh on 20091107 at 18:47 Reason: fixing tags 
20091107, 19:26  #5 
Just call me Henry
"David"
Sep 2007
Liverpool (GMT/BST)
37×163 Posts 
boy my brain wasnt thinking straight when i wrote that post!!!!

20091108, 01:17  #6 
May 2009
Dedham Massachusetts USA
3×281 Posts 
If you wanted to calculate the odds of losing the down driver, one can calculate the chance of 2*p occurring. However, there is also the case of
2*n^2*p where n is an odd number not divisible by p which can also escape. Therefore the odds of escape are better than the chance of getting 2*p with p = 1 mod 4. So the chance of an odd number being prime is the natural log of the odd part of the number and the chance of escape is therefore greater than 1/2 that. The chance of escaping with 2*p, p=1 mod 4 for a 100 digit number is 1 in 460.5 (ln (10^100)*2 = 460.5). I don't think the fact that we can have a square adds much. For example the chance of getting a square of 5 is only 1/25. Since we can't get a square of 2 or 3, the chance of a square part will be bigger than 1/25, but not a lot bigger. One can add 1/25 + 1/49+ 1/121 to get an aproximation of the chance of escape to be around 1/430 for a 100 digit number. Of course as the number gets lower, the chance of escape will increase. I was thinking one could do the calculation of figuring out the average number of steps it takes to lose a digit and do the calculation of the chance of getting from 200 to 100 digits assuming the steps were all equal the average. This would be a pretty good approximation of the chance a 200 digit sequence could escape before hitting 100 digits. At some point I may try to do the math for this. It would be interesting to know the answer. Last fiddled with by Greebley on 20091108 at 01:19 
20091108, 14:39  #7  
Just call me Henry
"David"
Sep 2007
Liverpool (GMT/BST)
178F_{16} Posts 
Quote:
Is this the formula you used when you included the possibility of squares? That gives me 1 in 431. Maxima simplifies it to: How do you get maxima to actually calculate ln(10^100)? I had to use google calc to get an approximation. 

20091108, 16:19  #8  
"Jacob"
Sep 2006
Brussels, Belgium
11101110001_{2} Posts 
Quote:
So all you can get is an approximation since ln(10) and 1 / log_{10}(e) are transcendental numbers. Jacob Last fiddled with by S485122 on 20091108 at 16:20 Reason: a 0 to many 

20091108, 17:05  #9  
"Gary"
May 2007
Overland Park, KS
2^{3}×7×211 Posts 
Quote:
Henryzz was right in the 1st post. How does this differ than the 1st post? lmao Am I losing my mind? 

20091108, 17:55  #10 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
11×389 Posts 

20091108, 18:06  #11 
Nov 2008
2×3^{3}×43 Posts 
ln(10^{100}) to 5 sig fig is 230.26.

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