20111216, 09:18  #1  
"Frank <^>"
Dec 2004
CDP Janesville
2·1,049 Posts 
Special for LaurV
Quote:
Code:
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 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. Last fiddled with by schickel on 20111216 at 09:20 Reason: Adding PS. 

20111216, 09:40  #2 
Romulan Interpreter
Jun 2011
Thailand
5×29×61 Posts 
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 23 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. Code:
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 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. 
20111216, 10:19  #3 
"Frank <^>"
Dec 2004
CDP Janesville
2·1,049 Posts 

20111220, 02:44  #4 
Romulan Interpreter
Jun 2011
Thailand
228D_{16} Posts 
181428 went back to the "plain" driver after two iterations.
Code:
181428 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). Code:
618480 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. Last fiddled with by LaurV on 20111220 at 03:26 
20111220, 05:55  #5 
Romulan Interpreter
Jun 2011
Thailand
5·29·61 Posts 
Done with 181428.
Code:
181428: T926. C121 = 2^3 * 3 * 5 * 67 * 89 * 25469 * 235510459 * C102 C102 cofactor was ecmed with "light" version of yafu only. Last fiddled with by LaurV on 20111220 at 05:56 
20111221, 03:59  #6 
Romulan Interpreter
Jun 2011
Thailand
5·29·61 Posts 
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.
Code:
865152 Last fiddled with by LaurV on 20111221 at 04:00 
20111221, 07:48  #7  
"Frank <^>"
Dec 2004
CDP Janesville
100000110010_{2} Posts 
Quote:
Quote:
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20111221, 08:08  #8  
Romulan Interpreter
Jun 2011
Thailand
5·29·61 Posts 
Quote:
Quote:
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. Last fiddled with by LaurV on 20111221 at 08:37 

20111221, 22:25  #9  
"Frank <^>"
Dec 2004
CDP Janesville
100000110010_{2} Posts 
Quote:


20111222, 03:51  #10 
Romulan Interpreter
Jun 2011
Thailand
5·29·61 Posts 
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 redefining and renaming 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 immediatefollowing 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*(2^41)), 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. Last fiddled with by LaurV on 20111222 at 04:08 
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