A new driver? (or type of driver?)

10metreh · 2010-02-15, 08:03

2^8 * 7 * 73 is a guide. According to the rules, it is class 4, so it isn't too hard to escape.
However, if you add in a 5, a 19 and a 37 to produce 2^8 * 5 * 7 * 19 * 37 * 73, then the 2^8 keeps the 7 and the 73 there, the 19 keeps the 5 there, the 37 keeps the 19 there and the 73 keeps the 37 there. The main difference is that the new primes raise the power of 2 of the sigma to 8, and other prime factors will raise it to 9 or more, meaning that it will keep the 2^8.
As far as I can see, you can only escape this structure when one of the factors in it is squared (like a driver that isn't 2^3 * 3 or the downdriver), but according to Clifford Stern's page (linked above), the guide is just the 2^8 * 7 * 73. The original definition of drivers and guides only allows for drivers formed from factors of the sigma of the power of 2, but clearly other primes can have a huge effect.

I apologise if this is incorrect.

TimSorbet · 2010-02-15, 12:00

A few experiments on your new "driver": (2^8*5*7*19*37*73 = 459818240)
http://factordb.com/search.php?se=1&...ange&fr=0&to=2 (37^2 at index 1; no 19, and so lost the driver, at index 2)
http://factordb.com/search.php?se=1&...nge&fr=0&to=10 (7^2 at index 5; 2^7, and so lost the driver, at index 6)
http://factordb.com/search.php?se=1&...ange&fr=0&to=2 (index 1=driver*3^2; 2^9, and so lost the driver, at index 2; it was lost without any of the driver's factors being squared, just that the non-driver cofactor was a square)
Doesn't seem to have much staying power.

10metreh · 2010-02-15, 14:06

Of course it's not like ordinary drivers in that it can keep the 2^8 while losing one of its factors, and it doesn't seem to stay for long, it's just the fact that it needs a squared factor to disappear that interests me.

Greebley · 2010-02-15, 15:57

The main difference is that a square will have a cascade effect in that all those dependent on the square term can be lost. So in MiniGeek's example the 37^2 lost the 19 which will eventually lose the 5 term and the 3 term (which wasn't listed but is based on the 5). Once the 19 is lost it is unlikely to get it back (1/19 every time we have a 37^2 term?)

Since the 37 would be squared every 37 iterations on average (I think), and keeping the 19 isn't very high if we do get a square, the long term stability is much lower than for a true driver. In the shorter term though, it is likely to last

2010-02-15, 08:03	#1
10metreh Nov 2008 2·3³·43 Posts	A new driver? (or type of driver?) 2^8 * 7 * 73 is a guide. According to the rules, it is class 4, so it isn't too hard to escape. However, if you add in a 5, a 19 and a 37 to produce 2^8 * 5 * 7 * 19 * 37 * 73, then the 2^8 keeps the 7 and the 73 there, the 19 keeps the 5 there, the 37 keeps the 19 there and the 73 keeps the 37 there. The main difference is that the new primes raise the power of 2 of the sigma to 8, and other prime factors will raise it to 9 or more, meaning that it will keep the 2^8. As far as I can see, you can only escape this structure when one of the factors in it is squared (like a driver that isn't 2^3 * 3 or the downdriver), but according to Clifford Stern's page (linked above), the guide is just the 2^8 * 7 * 73. The original definition of drivers and guides only allows for drivers formed from factors of the sigma of the power of 2, but clearly other primes can have a huge effect. I apologise if this is incorrect. Last fiddled with by 10metreh on 2010-02-15 at 08:04

2010-02-15, 12:00	#2
TimSorbet Account Deleted "Tim Sorbera" Aug 2006 San Antonio, TX USA 10267₈ Posts	A few experiments on your new "driver": (2^857193773 = 459818240) http://factordb.com/search.php?se=1&...ange&fr=0&to=2 (37^2 at index 1; no 19, and so lost the driver, at index 2) http://factordb.com/search.php?se=1&...nge&fr=0&to=10 (7^2 at index 5; 2^7, and so lost the driver, at index 6) http://factordb.com/search.php?se=1&...ange&fr=0&to=2 (index 1=driver3^2; 2^9, and so lost the driver, at index 2; it was lost without any of the driver's factors being squared, just that the non-driver cofactor was a square) Doesn't seem to have much staying power. Last fiddled with by TimSorbet on 2010-02-15 at 12:01

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2010-02-15, 14:06	#3
10metreh Nov 2008 2322₁₀ Posts	Of course it's not like ordinary drivers in that it can keep the 2^8 while losing one of its factors, and it doesn't seem to stay for long, it's just the fact that it needs a squared factor to disappear that interests me.

2010-02-15, 15:57	#4
Greebley May 2009 Dedham Massachusetts USA 3×281 Posts	The main difference is that a square will have a cascade effect in that all those dependent on the square term can be lost. So in MiniGeek's example the 37^2 lost the 19 which will eventually lose the 5 term and the 3 term (which wasn't listed but is based on the 5). Once the 19 is lost it is unlikely to get it back (1/19 every time we have a 37^2 term?) Since the 37 would be squared every 37 iterations on average (I think), and keeping the 19 isn't very high if we do get a square, the long term stability is much lower than for a true driver. In the shorter term though, it is likely to last