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[gd_barnes]

I am now done with 12456 and 101862. Statuses:

12456; i=1895; sz 106; C105; +60; 2*3
101862; i=1770; sz 109; C104; +64; 2*3^2

I'm giving up all these interesting sequences of late. In the last 3 days, I've now given up two with factors of only 2*3 and one with a factor of 2^2 that just dropped its factor of 3^2 just as I hit a hard C105.

Personally I don't find the 2*3 factorization particularly interesting because it is so tenacious, even if it is only slowly increasing. I have another one for 280686 that is an ever-increasing sequence that has never dropped 2*3 and it is at i=943!


I am posting any new reservation in the reservations thread.

[gd_barnes]

Originally Posted by 10metreh

Let me just add a note here. Although you can expect to have the 2^12 * 8191 driver for a very long time, I quote Guy and Selfridge from "What Drives an Aliquot Sequence?",

I'm sure that is a mathematically proven fact since there appears to be a specific condition for all perfect # factors.

Alas our computational capactiy forces even the most robust factor searchers with huge resources to abandon most or all sequences at a size of 150-170 somewhere. I'm guessing with a chance of 1 in the thousands of dropping the 2^12*8191 driver, it's one I would gladly give up once I hit a hard C100 or higher, even if it is considered "interesting".

BTW, correct me if I'm wrong but it doesn't have to be a perfect # or even a prime multiplier on top of the power of 2. I've only drawn this conclusion through observation. I believe a "tenacious driver" has to only be of the form:

2^n*[2^(n+1)-1]

2^(n+1)-1 does not even have to be prime. These would include:

2*3
2^2*7
2^3*15 = 2^3*3*5
2^4*31
2^5*63 = 2^5*3^2*7
2^6*127
2^7*255 = 2^7*3*5*17
2^8*511 = 2^8*7*73
2^9*1023 = 2^9*3*11*31
2^10*2047 = 2^10*23*89
2^11*4095 = 2^11*3^2*5*7*13
2^12*8191

Correct me if I'm wrong but the above would be quite tenacious drivers that would be hard to get rid of with their difficulty increasing as the power of 2 increases.

Can someone state what specific condition is required for each of the drivers above to be dropped? I see that 2*3 can be dropped if an index factors as 2^2*[P==(1 mod 4)]. I'd like to see the pattern of conditions that make drivers of the form 2^n*[2^(n+1)-1] go away.


Gary

[Greebley]

They are all guides. The drivers are the ones that have powers of two from p+1 for all the p that is close to the exponent of two.

For example, for 2^10*23*89 has 2^3 from 23 and 2 from 89 so it is short by 2^6. For higher ranges it is easily possible for the remaining numbers to have more than 2^6 among all the factors, but it is also not that hard to have 2^6 or less so the exponent changes.

An interesting thing to notice is that for larger perfect numbers, the chances of getting a square increases, but if you do get one the chances of escaping also increases for 2^12*8191^2*rest - the rest would have to have 13+ powers of two to keep the 2^12. 2^2*7^2 only needs two powers of 2 which is fairly likely.

The 43rd Mersenne prime would be very hard to get a square but when it does there would be millions of digits needed for the square.

[Mini-Geek]

Originally Posted by gd_barnes

I'm sure that is a mathematically proven fact since there appears to be a specific condition for all perfect # factors.

"What Drives an Aliquot Sequence?" outlines a proof that "certain prime factors of n [will never always] continue to appear to the same or higher powers in s(n)". If you haven't already, I recommend reading that paper. (you can get it from here)

Originally Posted by gd_barnes

2^n*[2^(n+1)-1]

I'd prefer if you used 2^(n-1)*(2^n-1), more similar to the common form shown for Mersenne primes, to keep with the convention.

Originally Posted by gd_barnes

2^3*15 = 2^3*3*5
2^5*63 = 2^5*3^2*7
2^7*255 = 2^7*3*5*17
2^9*1023 = 2^9*3*11*31

I'm not sure how much of this you may already be aware of:

2^3*3*5 is recognized as a driver.
2^5*3^2*7 is an alternate form of 2^5*3*7, and relies on it for stability.
"What Drives ..." claims 2^7*3^6*5*17*23*137*547*1093 is a fragile structure (notice the similarities to 2^7*3*5*17).
2^9*3*11*31 is recognized as a driver (albeit a rare one to encounter).
"What Drives ..." defines drivers and says (with a mathematical proof I can't say I understand) that the only drivers are 2, (the downdriver) 2^3*3, 2^3*3*5, 2^5*3*7, 2^9*3*11*31, and the even perfect numbers.

Is there any quantifiable way to measure how difficult a driver is to break free from? Perhaps somehow by listing every way it could be lost and calculating the probability of losing it on any one line?

Originally Posted by gd_barnes

Can someone state what specific condition is required for each of the drivers above to be dropped? I see that 2*3 can be dropped if an index factors as 2^2*[P==(1 mod 4)]. I'd like to see the pattern of conditions that make drivers of the form 2^n*[2^(n+1)-1] go away.


Gary

I don't know how to find specific requirements, (I'd be interested in hearing them, too ) but if a sequence with the 2*3 driver factors as 2^2*p, hasn't it already lost the 2*3 driver?

[gd_barnes]

Originally Posted by Batalov

My pet sequence had another downdriver which it lost and acquired 2*3 which it overnight recently lost to 2^2*3 and just now via 2^2*p_1mod4 acuired the downdriver again

Originally Posted by Mini-Geek

I don't know how to find specific requirements, (I'd be interested in hearing them, too ) but if a sequence with the 2*3 driver factors as 2^2*p, hasn't it already lost the 2*3 driver?

Oops, I misquoted what Batalov had said. See above. Ignore what I said about 2^2*[p==(1mod4)] causing the 2*3 driver to go away. He was referring to 2^2*3.

Therefore if anyone can give the condition(s) that cause the 2*3 driver to go away or even the more specific condition(s) that turn it into a downdriver, I'd be very interested in seeing them. I'd also be interestered in hearing the same that I gave for factors of the form 2^(n-1)*(2^n-1).


Gary

[10metreh]

Originally Posted by Mini-Geek

Is there any quantifiable way to measure how difficult a driver is to break free from? Perhaps somehow by listing every way it could be lost and calculating the probability of losing it on any one line?

Not quite - when you have the 2 * 3 driver, the chances are much less than 1/3 that the 3 will be squared on the next line. However, when the 3 is squared, it will tend to stay for quite some time (unless this is sequence 130396).

In order to work out the exact probabilities, you'd have to work out things like the chances that a number will split into two primes of the form 4n+1. Of course this depends on the size of the number, so you will have a better chance of escaping drivers low down, as you've probably noticed. But 2^6*127 and 2^12*8191 have the prime squared so rarely that the driver will drive them out of easy escape range before there's even been a chance.

[mataje]

435960: line 585, size 100, 2*3^2*7^2
445188: line 1112, size 100, 2^2*5*7
450054: line 1534, size 101, 2^3*3*5
454164: line 668, size 101, 2^2*3*7
457884: line 2344, size 101, 2^3*3*5
467994: line 957, size 100, 2^2*3^2

[Greebley]

As far as the conditions for 2*3, I believe they are that the 3 occurs to an even power, and the non-square part of the rest has to be 1 or a single prime congruent to 1 mod 4. If it is 1, you have 2 times a perfect square and the sequence becomes odd. Otherwise it is divisible by 2 but not 4. Then sigma and n are both = 2 mod 4 so sigma - n becomes divisible by 4 and you get 2 to a power greater than 1.

so for example 2*3^2*17^2*101 becomes 2^2*3^3*17*378 and the 2 is lost.

[10metreh]

Originally Posted by Greebley

As far as the conditions for 2*3, I believe they are that the 3 occurs to an even power, and the non-square part of the rest has to be 1 or a single prime congruent to 1 mod 4. If it is 1, you have 2 times a perfect square and the sequence becomes odd. Otherwise it is divisible by 2 but not 4. Then sigma and n are both = 2 mod 4 so sigma - n becomes divisible by 4 and you get 2 to a power greater than 1.

so for example 2*3^2*17^2*101 becomes 2^2*3^3*17*378 and the 2 is lost.

And to continue with 2^2 * 7, the 7 has to be raised to an even power, and the non-square part of the rest has to be 1, a single prime of the form 4n+1 or 8n+3, or two primes of the form 4n+1.

I could extend to 2^4 * 31, that's a bit more complicated.

(And don't make me think about completely explaining 2^6*127 or 2^12*8191 )

[Mini-Geek]

Why are so many driver breakings reliant on primes with p==1 mod 4? (the downdriver, at least two of the perfect number drivers, etc.)

[Greebley]

Because p+1 has only 1 power of 2 when p is 1 mod 4. Since drivers only switch when powers of two change and powers of two only change when the power of two is low enough, the numbers that have the fewest powers of two are what makes things change.

The p+1 comes from the calculation of sigma whenever p isn't a square.