Special for LaurV

[schickel]

Originally Posted by LaurV

Ouch! One of the worst - read most persistent and fast increasing - drivers.... (2^3*(2^4-1))... If you can get rid of it, please let us know, I am yet to see how this driver can be broken... In fact, I am dying to see that, see my posts about 585000 in the reservation thread. Usually it holds for hundreds of terms....

Who said "Fear the 2^3*3*5" was damn right!

In light of your interest, here are sequences with 2^3 * 3 * 5 that might manage an escape:

982290   500. sz 117 2^3 * 3^2 * 5^2
613068   543. sz 116 2^3 * 3^2 * 5^2
136780   549. sz 116 2^3 * 3^6 * 5^2
773070   696. sz 117 2^3 * 3^2 * 5^2
722904   855. sz 114 2^3 * 3^2 * 5^2
181428   902. sz 112 2^3 * 3^2 * 5^2
189140   963. sz 115 2^3 * 3^2 * 5^2
696204  1055. sz 117 2^3 * 3^2 * 5^2
225900  1085. sz 113 2^3 * 3^2 * 5^2
612960  1352. sz 120 2^3 * 3^2 * 5^2
865152  1434. sz 114 2^3 * 3^4 * 5^2
618480  1735. sz 112 2^3 * 3^2 * 5^2
814848  1938. sz 115 2^3 * 3^2 * 5^2

The 3 shortest ones will have had the the driver from near the start of their runs.....

The best thing to do is to check the status in the DB and see if they've had the full round of ECM from the "Scan" buttons, then run some local ECM to see if anything small shakes out.

[LaurV]

Don't touch them! Take off your hand!
I will draw my scimitar and point it to their necks when after I arrive home (about 2-3 hours). What I can do right now is to add the cofactors, to have a starting point, and this I will do at once using factorDB.

982290   500. C117 = 2^3 * 3^2 * 5^2 * C114
613068   543. C116 = 2^3 * 3^2 * 5^2 * C112
136780   549. C116 = 2^3 * 3^6 * 5^2 * C111
773070   696. C117 = 2^3 * 3^2 * 5^2 * C114
722904   855. C114 = 2^3 * 3^2 * 5^2 * C111
181428   902. C112 = 2^3 * 3^2 * 5^2 * C109
189140   963. C115 = 2^3 * 3^2 * 5^2 * C112
696204  1055. C117 = 2^3 * 3^2 * 5^2 * C114
225900  1085. C113 = 2^3 * 3^2 * 5^2 * C110
612960  1352. C120 = 2^3 * 3^2 * 5^2 * C117
865152  1434. C114 = 2^3 * 3^4 * 5^2 * C110
618480  1735. C112 = 2^3 * 3^2 * 5^2 * C109
814848  1938. C115 = 2^3 * 3^2 * 5^2 * C112

The size of the cofactors seems all "tangible". Hope none of them are reserved already...

865152 and 136780 look appealing because of the power of 3, but really I don't know if this helps. According with my modular calculus, this driver will not get lost unless the power of 2 increase to something even, as 4 or 6, otherwise the sum will almost always be divisible by 3 and by 5. And to increase the power of two, I got lost in the (mod 4), (mod 3) and (mod 7) calculus, without a result.

[schickel]

Originally Posted by LaurV

Hope none of them are reserved already...

I didn't check when I posted, but a quick check now says that none of them are reserved.

[LaurV]

181428 went back to the "plain" driver after two iterations.

181428   902. C112 = 2^3 * 3^2 * 5^2 * C109
181428   925. C120 = 2^3 * 3 * 5 * 143807 * C113

I added 23 terms to it, and 8 digits, the C113 is in process (nfs almost finished when I reached this computer today, so I would let it finish). I will stop (and unreserve it) when I reach the next nfs-able cofactor.

618480 switched immediately to 2^3*3^3*5, added 9 digits and 28 terms, almost all of them 2^3*3^3*5, and is now at term 1763, where it switched back to 2^3*3^2*5^2. Still going on it, and I will not reach that computer within today (therefore a couple of terms will be added over the one shown here).

618480  1735. C112 = 2^3 * 3^2 * 5^2 * C109
618480  1763. C123 = 2^3 * 3^2 * 5^2 * 7 * 53^3 * 379 * 143629 * C106

Wondering if the high power of 53 could mean something, or if it could help. It seems that 53 appears quite regularly in the sequence.

All the other sequences I will continue until the driver is lost or it switches back to the "plain" form 2^3*3*5, or until they get to complicate for me to factor them.

[LaurV]

Done with 181428.

181428: T926. C121 = 2^3 * 3 * 5 * 67 * 89 * 25469 * 235510459 * C102

I'll let the factorDB's elves to raise the cofactor few digits. Or anyone else could like to try? :P
C102 cofactor was ecm-ed with "light" version of yafu only.

[LaurV]

865152 managed an escape and got the downdriver after few (6) iterations, it seems that I was right assuming the 3^4 was somehow important.

865152  1434. C114 = 2^3 * 3^4 * 5^2 * C110
865152  1440. C116 = 2 * C115

If someone wants the glory of getting an eventual termination , I am willing to pass him the sequence after current 865152:T1440. C116 = 2 * C115 will finish the nfs (about 60% done, few hours left). Just say so here, in this thread. If not, I will continue with it, until a possible termination or evaporation of the downdriver will occur. My interest is to see how this driver escapes. Of course, I would like to ride the downdriver myself, this is quite exciting, but my resources are 120% busy this time, and all this work in this thread is kinda "overload" due to my curiosity, I could not stop taking it, and it affects negative my output to GPU-2-72 project.

[schickel]

Originally Posted by LaurV

865152 managed an escape and got the downdriver after few (6) iterations,

Incredible! That's more than I was expecting; the low-powers-of-2 drivers are really hard to escape.

it seems that I was right assuming the 3^4 was somehow important.

I don't know how much reading you have done, but have you read Clifford Stern's analysis page? The important thing is that the 3 has an even power....

My interest is to see how this driver escapes. Of course, I would like to ride the downdriver myself, this is quite exciting, but my resources are 120% busy this time, and all this work in this thread is kinda "overload" due to my curiosity, I could not stop taking it, and it affects negative my output to GPU-2-72 project.

I apologize; I didn't mean to distract you from your preferred work!

[LaurV]

Originally Posted by schickel

Incredible! That's more than I was expecting; the low-powers-of-2 drivers are really hard to escape.I don't know how much reading you have done, but have you read Clifford Stern's analysis page? The important thing is that the 3 has an even power....

I have read that, and also did a lot of calculus by myself, especially for this driver (after my 585000 locked itself to it), less for the other, but still (for example 3^4 here is more important then 3^6, and 3^6 is more important then 3^2. But as long as 2 will not get an even power, the driver can't escape). I also "touched" powers (sequences starting with powers) and odd numbers (sequences without 2). Unfortunately, as I said in a different thread, I got lost in the modular stuff and could not come with a consistent result, my theoretical knowledge is not much deeper then basic calculus. My belief, in spite of the most recent results against the conjecture, is that all aliquots will end in a prime or a cycle sooner or later (perfect numbers = cycles of length 1).

I apologize; I didn't mean to distract you from your preferred work!

No need to. You did not force me.

P.s please consider I reserved all these sequences and add them to the main reservation thread. Including 181428 (which I said above that I am done with it, but as long as no one wants it, I will keep it on the list). However, you should be aware that the progress will be slow, I play with them "sequentially", I have not enough cores free to run all in the same time.

[schickel]

Originally Posted by LaurV

I have read that, and also did a lot of calculus by myself, especially for this driver (after my 585000 locked itself to it), less for the other, but still (for example 3^4 here is more important then 3^6, and 3^6 is more important then 3^2. But as long as 2 will not get an even power, the driver can't escape).

You have a circular argument there. If you have 2^3 * 3 * 5 as the driver, you're saying you can't escape the driver unless the 2 gets an even power. The thing is, though, if the power on the 2 changes, the driver has already been escaped. The driver is defined by the power of 2: if you have 2^3 plus 3 and 5, you have the 2^3 * 3 * 5 driver (or the guide form if either 3 or 5 has a power >1). If you have 2^4 plus 3 and 5, it's not the 2^3 * 3 * 5 driver. (It could be the 2^4 * 31 driver, but the 3 and/or 5 aren't locked in like they with 2^3.) What Clifford's analysis says is that if the 3 and/or the 5 are raised to an even power, the result is a guide that is in a different class than the "pure" driver, which gives you a chance to escape the driver.

[LaurV]

Originally Posted by schickel

You have a circular argument there.

Well, you are right. As usually, I have a different "perception"... (as RDS would say, "the first sign of a crank is that he starts re-defining and re-naming known terms" -- obviously because he did not know about the existence of them, that is he did not spent any effort to learn the theory, this was my addition to what RDS said -- "and invents his own new names/terms")

For some drivers, it could be possible to "raise/decrease the power of 2" to something even, and immediately after, the power of two be back to odd. Or viceversa. Take for example 2^4*31, it is possible for the power of 2 to oscillate between 4 and 5 (and even 7 or 8) but the driver will not get lost, it will eventually come back to the "plain" form sooner or later. I will not consider any of this being an "escape". "Escape" means for me that the driver is gone, and no immediate chances for it to come back in the immediate-following steps.

Some combination of factors will have higher chances to "persist" along the sequence then other combinations. These are all called "guides". You can split them in classes in many different ways. Each guide could change the "bias" of the sequence to "up" or to "down". That is, the next term will not necessarily be higher (or respective lower), but after a couple of terms the sequence will have better chances to increase (or respective to decrease). When the guide is of the form 2^n*(2^(n+1)-1), then I call it "driver". If a driver has better chances to decrease the sequence, then is a downdriver. I can prove that any "guide" is a "partial" form of a driver. That is, a driver form which some terms a missing. That is easy, and not very useful OTOH. The drivers could then be something like a "perfect guide" or "complete guide". It could be the "plain" form, as 2^3*3*5 (from 2^3*(2^4-1)), or it could have variations in powers of 3, 5, AND 2. For example it could lose the 5 at all, then get it back at the next step. "Fives and threes come and go". It could go to 2^4*3^2*5^x*a*b with x=1 or x=3, a and b primes, and then come back to 2^3*3*5 (this is really possible, depending on the properties of a and b, their modularity to 4, 3, and 7). I won't consider that an "escape".

Of course, you can classify them in many ways. My classification come from the simple observation: if you have a perfect number, the sequence stays the same. Then we take out any prime factor of the perfect number and substitute it with another random prime. If this prime is higher then the factor we took out, the sequence decreases more (or increases slower). If it is smaller, the sequence increases more (or decreases slower). Then we generalize, somehow: if you take out any factor from any number, and substitute with a higher one, its sigma increase less (or decrease more). And viceversa. From here, it is just modular computing, taking a guide times a prime, a guide times two primes, a guide times 3 primes, playing with different guides, and with the modularity of the primes regarding 3,4,5,7, etc.

Take for example the sequence starting with 18486235136 (=2^12*8191*p*q, with p=19 and q=29, different parities mod 4). There are thousands of 3's and 5's (and 7's and 11's, etc) that come and go, but the only one which stays stable hundreds(thousands??) of terms is 2^12*8191. From term 300++ they even come together (as 2^12*3*5*...) but they will not hold. Interesting that aliqueit ("known theory") recognize that driver as "large power of two", but if we replace 8191 in any term with almost any other prime, the powers of two will be gone in a blink of an eye. Always! Powers of two are not drivers. So, in a word, what you call driver, I call "plain driver". What I call driver, is in fact a larger class, including plain driver and a partition of its associated guides. For me 2^4*3^2*5^3 is the same driver as 2^3*3*5, but not in its plain form, AS LONG AS the 31 is missing. If 31 is there, then the driver is 2^4*31, in its plain form, and all the 3's and 5's will be almost sure gone in the next iterations.

[schickel]

You're the one that brought up crankery.....but your definition of a driver is different from the definition in the Guy and Selfridge article; maybe we need to invent a new term: D_{[COLOR="Blue"]LV[/COLOR]} for LaurV's Driver.