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 2010-02-02, 21:04 #1 fivemack (loop (#_fork))     Feb 2006 Cambridge, England 645410 Posts What do I have to do to break free? 9336 iteration 1037 splits as 1119687698...<144> = 2^6 · 7 · 127^2 · 12577 · 11466743 · 86572333 · 1241124045...<118> but nonetheless iteration 1038 has the 127^1 still there, still mocking me with its evil set of seven set bits. What is the probability of driver escape at this level? (similarly, I've had 31^2 at iterations 3916, 3930 and 3962 of 6822, but 31^1 returns the very next line) Last fiddled with by fivemack on 2010-02-02 at 21:06
2010-02-03, 07:24   #2
10metreh

Nov 2008

2×33×43 Posts

Quote:
 Originally Posted by fivemack 9336 iteration 1037 splits as 1119687698...<144> = 2^6 · 7 · 127^2 · 12577 · 11466743 · 86572333 · 1241124045...<118> but nonetheless iteration 1038 has the 127^1 still there, still mocking me with its evil set of seven set bits. What is the probability of driver escape at this level? (similarly, I've had 31^2 at iterations 3916, 3930 and 3962 of 6822, but 31^1 returns the very next line)
The 2s count of a prime number is the exponent of the highest power of 2 that divides that prime + 1. To escape 2^6 * 127, firstly, the 127 must be raised to an even power. Also, the sum of the 2s counts of all the other prime factors must be no greater than 6.

 2010-02-04, 15:45 #3 Greebley     May 2009 Dedham Massachusetts USA 3×281 Posts The 7 term is the evil one. It gives 3 factors of 2 and will stick around as long as some term is 6 mod 7. I would like to see how many factors of two the 'large terms' (ones that don't stick around) add for the different number of digits. It might be possible to calculate that - or at least a good estimate. The average is higher than 3 however which means that with the 7 term you are unlikely to escape. Last fiddled with by Greebley on 2010-02-04 at 15:46
2010-02-05, 16:45   #4
fivemack
(loop (#_fork))

Feb 2006
Cambridge, England

2·7·461 Posts

Quote:
 Originally Posted by Greebley The 7 term is the evil one. It gives 3 factors of 2 and will stick around as long as some term is 6 mod 7. I would like to see how many factors of two the 'large terms' (ones that don't stick around) add for the different number of digits. It might be possible to calculate that - or at least a good estimate. The average is higher than 3 however which means that with the 7 term you are unlikely to escape.
Well, val_2(p+1) has mean 2, the mean number of prime factors for a number around N is (theorem of Erdos) loglog N, and I think is expected to be Poisson-distributed with mean loglog N; so around 10^100 you'd expect there to be about 5.5 prime factors; 2 and 127 account for two of those, leaving about 3.5 big ones each contributing about two factors of two. So the numbers get big, and escape is unlikely.

2010-02-05, 18:15   #5
CRGreathouse

Aug 2006

598810 Posts

Quote:
 Originally Posted by fivemack Well, val_2(p+1) has mean 2, the mean number of prime factors for a number around N is (theorem of Erdos) loglog N, and I think is expected to be Poisson-distributed with mean loglog N; so around 10^100 you'd expect there to be about 5.5 prime factors; 2 and 127 account for two of those, leaving about 3.5 big ones each contributing about two factors of two. So the numbers get big, and escape is unlikely.
I would expect more like 5 ≈ log(log(1e100 / 254)) - 1/2 - 1/127 other prime factors, if they behave like random multiples of 254 near 1e100.

 2010-03-07, 07:41 #6 schickel     "Frank <^>" Dec 2004 CDP Janesville 22×32×59 Posts I feel your pain*.... From 20916: Code:  670 . c117 = 2^6 * 3 * 11 * 13 * 127^2 * 84584688662069 * 303457268457331 * 23305983260885734320137249 * 3127792565339118661368476021978704179191553219477805861 671 . c118 = 2^6 * 3 * 11 * 83 * 127 * 839 * c107 And not only did it gain a digit that line, that cursed 3 has been turning up the heat (16 lines so far)! * Line stolen from Bill Clinton [inside joke for the USAians here]
 2010-04-08, 13:08 #7 TimSorbet Account Deleted     "Tim Sorbera" Aug 2006 San Antonio, TX USA 10B716 Posts How do I lose 2^5 * 7 (not a listed driver or guide, but it seems like a guide to me)? It's taken over sequence 652296 for 75 lines (going up 11 digits) so far. Last fiddled with by TimSorbet on 2010-04-08 at 13:08
2010-04-08, 18:11   #8
TimSorbet
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

11×389 Posts

Quote:
 Originally Posted by Mini-Geek How do I lose 2^5 * 7 (not a listed driver or guide, but it seems like a guide to me)? It's taken over sequence 652296 for 75 lines (going up 11 digits) so far.
I lost it here: http://factordb.com/search.php?se=1&...r=1446&to=1448
At line 1447, the 2^5 switched to 2^3, but it still had 7^2 as a factor. Line 1448 lost the 7 as well, so now the only guide is 2^3, so this big sequence is now in a downward slope.
Still leaves the question of how 2^5 * 7 (or 7 to any other power) can be broken directly? Or can it?

Last fiddled with by TimSorbet on 2010-04-08 at 18:11

 2010-04-08, 18:47 #9 10metreh     Nov 2008 44228 Posts It is a listed guide: http://www.lafn.org/~ax810/analysis.htm

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