Subproject #3: 700k-800k sequences to 100 digits - Page 21

rodac · 2009-11-17, 20:15

Quote:

Originally Posted by 10metreh

That (2^3 * 3) is a driver, but it's the easiest to escape of all drivers, especially while the 3 is squared.

ok I thought that 2^3 * 3 was a guide...

---

Done with 723120
index 1247, 104 digits, 2^2 * 3 * 7

10metreh · 2009-11-17, 20:42

Done with

Code:

728640 3554. 2^4 * 3 * 5 * 31 c94 sz 101

This one went down from 92 to 11 digits (took several downdriver runs), but then it acquired 2^3 * 3 at 31 digits, that soon changed directly into 2 * 3, and that changed directly into 2^3 * 3 * 5, which changed directly into 2^4 * 31 at 95 digits. Life is sad sometimes.

Greebley · 2009-11-17, 21:37

Quote:

Originally Posted by 10metreh

That (2^3 * 3) is a driver, but it's the easiest to escape of all drivers, especially while the 3 is squared.

I don't think it is a driver unless it has a factor of 5 and the exponent of 3 is one - so specifically 2^3*3^1*5

Anything else isn't fully supported - the powers of two from the 'one more than the odd terms' must be greater or equal to the power of two.

So for example 2^3*3 has only two twos from 3+1 which is less than the 3 of 2^3 so it is not a driver.

Another way of stating it is that it cannot change the power of 2 unless one of the odd terms is a power higher than 1. So [driver]*p where p = 1 mod 4 doesn't escape.

Mini-Geek · 2009-11-17, 22:25

Quote:

Originally Posted by Greebley

I don't think it is a driver unless it has a factor of 5 and the exponent of 3 is one - so specifically 2^3*3^1*5
...

http://www.mersennewiki.org/index.php/Aliquot_Sequences
2^3 * 3 is a driver (an easy-to-break and slow-growing one), as well as 2^3 * 3 * 5 (a tough-to-break and fast-growing one).

Buzzo · 2009-11-18, 02:22

done with 720912
reserving 723432

Batalov · 2009-11-18, 05:42

Done with 725052 729792 727596 and 729660

Greebley · 2009-11-18, 14:23

Ok, you are right. I looked at the Analysis and Class 1 (ones that can escape without a square term on the odd number if they get [driver]*p where p is 1 mod 4) is called a driver so that 2 can be included in the drivers. This has the side effect of including 2^3*3 as a driver as well.

I personally don't consider 2^3*3^2*5 to be a driver because this is class 2 and the exponent of 3 can't be lowered to 1 unless 5 is square. 2^5*3^2*7 can never go to 3^1 without changing the power of 2 so this one remains class 2. The analysis doesn't specifically mention these cases though so I am not sure if they are officially not drivers.

-----------
Done with 721364, 114 digits, 2^3*3*5*7

10metreh · 2009-11-18, 15:45

Quote:

Originally Posted by Greebley

Ok, you are right. I looked at the Analysis and Class 1 (ones that can escape without a square term on the odd number if they get [driver]*p where p is 1 mod 4) is called a driver so that 2 can be included in the drivers. This has the side effect of including 2^3*3 as a driver as well.

Actually the definition of a driver is that, for $2^an$ , $n\mid\sigma(2^a)$ and $2^{a-1}\mid\sigma(n)$ .

Buzzo · 2009-11-18, 16:07

done with 723432
reserving 723480

Greebley · 2009-11-18, 18:07

Quote:

Originally Posted by 10metreh

Actually the definition of a driver is that, for $2^an$ , $n\mid\sigma(2^a)$ and $2^{a-1}\mid\sigma(n)$ .

Actually that is the same thing, just stated more technically. The -1 in a-1 of the second expression is why class 1 is counted. In fact:
$2^{a-<class>}\mid\sigma(n)$

10metreh · 2009-11-18, 18:41

Reserving 726354

2009-11-17, 20:42	#222
10metreh Nov 2008 2·3³·43 Posts	Done with Code: 728640 3554. 2^4 * 3 * 5 * 31 c94 sz 101 This one went down from 92 to 11 digits (took several downdriver runs), but then it acquired 2^3 * 3 at 31 digits, that soon changed directly into 2 * 3, and that changed directly into 2^3 * 3 * 5, which changed directly into 2^4 * 31 at 95 digits. Life is sad sometimes.

2009-11-18, 14:23	#227
Greebley May 2009 Dedham Massachusetts USA 843₁₀ Posts	Ok, you are right. I looked at the Analysis and Class 1 (ones that can escape without a square term on the odd number if they get [driver]p where p is 1 mod 4) is called a driver so that 2 can be included in the drivers. This has the side effect of including 2^33 as a driver as well. I personally don't consider 2^33^25 to be a driver because this is class 2 and the exponent of 3 can't be lowered to 1 unless 5 is square. 2^53^27 can never go to 3^1 without changing the power of 2 so this one remains class 2. The analysis doesn't specifically mention these cases though so I am not sure if they are officially not drivers. ----------- Done with 721364, 114 digits, 2^3357 Last fiddled with by Greebley on 2009-11-18 at 14:25*

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2009-11-18, 02:22	#225
Buzzo May 2008 3·37 Posts	done with 720912 reserving 723432

2009-11-18, 05:42	#226
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 22431₈ Posts	Done with 725052 729792 727596 and 729660

2009-11-18, 16:07	#229
Buzzo May 2008 111₁₀ Posts	done with 723432 reserving 723480

2009-11-18, 18:41	#231
10metreh Nov 2008 2·3³·43 Posts	Reserving 726354