Losing downdriver - Page 3

Greebley · 2009-11-15, 19:27

Actually the article you quoted just above has an interesting point. Since we are asking if an ODD number is prime, then the chances are twice as high as a random number which could be even.

I therefore think the above posts might be incorrect in their estimates by a factor of 2 and the chance of prime for a 100 digit odd number is 1/115 and the chance of gaining the down driver is therefore 1/230.

Does that make sense?

henryzz · 2009-11-15, 20:22

Quote:

Originally Posted by Greebley

Actually the article you quoted just above has an interesting point. Since we are asking if an ODD number is prime, then the chances are twice as high as a random number which could be even.

I therefore think the above posts might be incorrect in their estimates by a factor of 2 and the chance of prime for a 100 digit odd number is 1/115 and the chance of gaining the down driver is therefore 1/230.

Does that make sense?

it does to me
the same thing had crossed my mind earlier

can anyone say whether my guess at the formula when including trialfactoring is correct or not?

henryzz · 2010-12-22, 13:47

Possible driver/guide acquisitions:

Code:

Driver       When p is
2^2 * 7      25 mod 56
2^3 * 5      37 mod 80
2^4 * 31     493 mod 992
2^5 * 7      221 mod 448
2^6 * 127    8125 mod 16256

Drivers and guides containing a 3 can't be obtained since 3 divides $\sigma(2)$ .
The remaining factors of the next iteration can also be worked out. For example:
The 2^2 * 7 driver is generated from 2*p by p = 56n+25. The remaining factors of the next iteration are the factors of 2n+1.

science_man_88 · 2011-05-11, 15:43

Quote:

Originally Posted by henryzz

Possible driver/guide acquisitions:

Code:

Driver       When p is
2^2 * 7      25 mod 56
2^3 * 5      37 mod 80
2^4 * 31     493 mod 992
2^5 * 7      221 mod 448
2^6 * 127    8125 mod 16256

Drivers and guides containing a 3 can't be obtained since 3 divides $\sigma(2)$ .
The remaining factors of the next iteration can also be worked out. For example:
The 2^2 * 7 driver is generated from 2*p by p = 56n+25. The remaining factors of the next iteration are the factors of 2n+1.

though I've already PM'd this, is it always that $p\equiv {<driver>-3} \text{ mod } {2 \time {<driver>}}$

henryzz · 2011-05-12, 15:53

Quote:

Originally Posted by science_man_88

though I've already PM'd this, is it always that $p\equiv {<driver>-3} \text{ mod } {2 \time {<driver>}}$

Sorry I misunderstood your pm and meant to go back to thinking about it later.
It seems that that is the case but I can't see why. Can anyone produce a proof?
It works the same for 2*5(not a driver or guide). 2*5 requires 7 mod 20. Even 2*5*7 follows the rule 67 mod 140 so it isn't just one factor other than 2 in the driver.

henryzz · 2011-05-12, 20:54

If the starting iteration is 2*p then the second iteration is 2p+3 which means than whatever we want the second iteration to be divisible by must divide 2p+3. which is what causes the noticed effect. Working backwards from any iteration with a possible downdriver on the previous iteration to the value of p for that iteration is p=((current iteration)-3)/2. Quite simple really when I thought for a while.

Batalov · 2011-05-12, 23:48

Quote:

Originally Posted by henryzz

If the starting iteration is 2*p then the second iteration is 2p+3 which means ...

If the starting iteration is 2*p, then the second iteration is p+3

2010-12-22, 13:47	#25
henryzz Just call me Henry "David" Sep 2007 Cambridge (GMT/BST) 5886₁₀ Posts	Possible driver/guide acquisitions: Code: Driver When p is 2^2 * 7 25 mod 56 2^3 * 5 37 mod 80 2^4 * 31 493 mod 992 2^5 * 7 221 mod 448 2^6 * 127 8125 mod 16256 Drivers and guides containing a 3 can't be obtained since 3 divides $\sigma(2)$ . The remaining factors of the next iteration can also be worked out. For example: The 2^2 * 7 driver is generated from 2*p by p = 56n+25. The remaining factors of the next iteration are the factors of 2n+1.

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2009-11-15, 19:27	#23
Greebley May 2009 Dedham Massachusetts USA 3×281 Posts	Actually the article you quoted just above has an interesting point. Since we are asking if an ODD number is prime, then the chances are twice as high as a random number which could be even. I therefore think the above posts might be incorrect in their estimates by a factor of 2 and the chance of prime for a 100 digit odd number is 1/115 and the chance of gaining the down driver is therefore 1/230. Does that make sense?

2011-05-12, 20:54	#28
henryzz Just call me Henry "David" Sep 2007 Cambridge (GMT/BST) 13376₈ Posts	If the starting iteration is 2*p then the second iteration is 2p+3 which means than whatever we want the second iteration to be divisible by must divide 2p+3. which is what causes the noticed effect. Working backwards from any iteration with a possible downdriver on the previous iteration to the value of p for that iteration is p=((current iteration)-3)/2. Quite simple really when I thought for a while.