20100202, 21:04  #1 
(loop (#_fork))
Feb 2006
Cambridge, England
6,379 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 20100202 at 21:06 
20100203, 07:24  #2  
Nov 2008
2·3^{3}·43 Posts 
Quote:


20100204, 15:45  #3 
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 20100204 at 15:46 
20100205, 16:45  #4  
(loop (#_fork))
Feb 2006
Cambridge, England
1100011101011_{2} Posts 
Quote:


20100205, 18:15  #5  
Aug 2006
2×11×271 Posts 
Quote:


20100307, 07:41  #6 
"Frank <^>"
Dec 2004
CDP Janesville
4112_{8} 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 * Line stolen from Bill Clinton [inside joke for the USAians here] 
20100408, 13:08  #7 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
4267_{10} 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 MiniGeek on 20100408 at 13:08 
20100408, 18:11  #8  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17·251 Posts 
Quote:
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 MiniGeek on 20100408 at 18:11 

20100408, 18:47  #9 
Nov 2008
2×3^{3}×43 Posts 
It is a listed guide: http://www.lafn.org/~ax810/analysis.htm

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