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Old 2009-11-07, 17:04   #1
henryzz
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Default 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.
25% If p = 5 mod 8 then the power of 2 is 3 or more:
12.5% If p = 5 mod 16 then the power of 2 is 3.
12.5% If p = 13 mod 16 then the power of 2 is 4 or more:
It continues like this infinitely.
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 2009-11-08 at 14:41 Reason: adding a little comment
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Old 2009-11-07, 17:18   #2
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Quote:
Originally Posted by henryzz View Post
50% If p = 1 mod 4 then we still keep the downdriver.
50% If p = 3 mod 4 then:
25% If p = 3 mod 8 then the power of 2 is 2.
25% If p = 6 mod 8 then:
12.5% If p = 6 mod 16 then the power of 2 is 2.
12.5% If p = 12 mod 16 then:
It continues like this infinitely.
Is the "power of 2 is 2" part supposed to increment at some point? Otherwise it would clearly mean simply:
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 Mini-Geek on 2009-11-07 at 17:19
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Old 2009-11-07, 18:17   #3
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Quote:
Originally Posted by Mini-Geek View Post
Is the "power of 2 is 2" part supposed to increment at some point? Otherwise it would clearly mean simply:
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.
looks like i got something badly wrong
i will correct it is a bit when i have time
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Old 2009-11-07, 18:40   #4
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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.
25% If p = 5 mod 8 then the power of 2 is 3 or more:
12.5% If p = 5 mod 16 then the power of 2 is 3.
12.5% If p = 13 mod 16 then the power of 2 is 4 or more:

Last fiddled with by 10metreh on 2009-11-07 at 18:47 Reason: fixing tags
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Old 2009-11-07, 19:26   #5
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boy my brain wasnt thinking straight when i wrote that post!!!!
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Old 2009-11-08, 01:17   #6
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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 2009-11-08 at 01:19
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Old 2009-11-08, 14:39   #7
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Quote:
Originally Posted by Greebley View Post
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.
Brilliant!! I love study like this.
Is this the formula you used when you included the possibility of squares?
(ln (10^{100})-{1\over1/25 + 1/49+ 1/121})*2 That gives me 1 in 431.
Maxima simplifies it to:
$$\frac{20358\,ln\left( 10^{100}\right) -296450}{10179}$$

How do you get maxima to actually calculate ln(10^100)? I had to use google calc to get an approximation.
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Old 2009-11-08, 16:19   #8
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Quote:
Originally Posted by henryzz View Post
How do you get maxima to actually calculate ln(10^100)? I had to use google calc to get an approximation.
ln(10100) = 100 * ln(10) = 100 / log10(e)

So all you can get is an approximation since ln(10) and 1 / log10(e) are transcendental numbers.

Jacob

Last fiddled with by S485122 on 2009-11-08 at 16:20 Reason: a 0 to many
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Old 2009-11-08, 17:05   #9
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Quote:
Originally Posted by 10metreh View Post
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.
25% If p = 5 mod 8 then the power of 2 is 3 or more:
12.5% If p = 5 mod 16 then the power of 2 is 3.
12.5% If p = 13 mod 16 then the power of 2 is 4 or more:

Henryzz was right in the 1st post. How does this differ than the 1st post? lmao Am I losing my mind?
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Old 2009-11-08, 17:55   #10
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Quote:
Originally Posted by gd_barnes View Post
Henryzz was right in the 1st post. How does this differ than the 1st post? lmao Am I losing my mind?
It was edited to match 10metreh's. The original, and very wrong, version is quoted in my earlier reply.
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Old 2009-11-08, 18:06   #11
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ln(10100) to 5 sig fig is 230.26.
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