20120402, 14:06  #1 
"Daniel Jackson"
May 2011
14285714285714285714
599 Posts 
Sequence 4198862272
I noticed this sequence has been stuck at a C110 since Valentines Day:
http://www.factordb.com/sequences.ph...20&fr=0&to=100 C110= Code:
51709575528507073763595405281798526430796959961820579558786286970151157498771359600065634235127123647562961971 Last fiddled with by Stargate38 on 02 Apr 12 at 07:10 Last fiddled with by schickel on 20120404 at 08:40 Reason: Adding [code] tags 
20120402, 15:34  #2 
(loop (#_fork))
Feb 2006
Cambridge, England
6,353 Posts 
This is me with my supermoderator hat on.
I would strongly appreciate it if you stopped sending messages complaining that nobody has yet done a straightforward calculation (it will take you about three days counting the time that it takes you to learn how to use gnfs) that you (and, for this sequence, only you) are interested in. If you continue doing so then I will consider transforming you for some mediumlength period into some silent amphibianlike object. 
20120402, 15:56  #3 
Apr 2010
Over the rainbow
2·3·5·79 Posts 
i will avance it a few iteration in the hope it will satisfy him

20120402, 16:46  #4  
Jun 2003
1001000111011_{2} Posts 
Quote:
If you want to see this sequence extended, you'll have to DIY. Last fiddled with by axn on 20120402 at 16:47 

20120402, 19:17  #5 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
1C35_{16} Posts 
Stargate, as axn said, do it yourself. With regards to learning how to do gnfs, you don't have to: all you need is YAFU. You can type "yafu 'factor(<number here>)'" and it will automatically throw some ECM at it (among other methods) for about 10 minutes, and then move on to SIQS or GNFS automatically and complete them without input from the user. PM me if you want more help (but I highly recommend just downloading YAFU and reading README, that should get you all you need to know).
Last fiddled with by Dubslow on 20120402 at 19:17 Reason: s/others/other methods/ 
20120402, 20:20  #6 
Apr 2010
Over the rainbow
2·3·5·79 Posts 
There, i advanced it by 10 line, you have now a C112(300 curves at 1e6), halfhearteadly ecm'ed.. now go play with your rage and your demands somewhere else... it took me 7 hours.
Now i ll go back to test R468 (n @72214) and R207 (n @24820) Last fiddled with by firejuggler on 20120402 at 20:22 
20120403, 14:30  #7 
"Daniel Jackson"
May 2011
14285714285714285714
599 Posts 
What are the criteria for losing the 2^3*3*5 driver? Does it have to be 2^3*3^2*5^2 or 2^3*3^3*5^3?
Look how straight the graph has gotten: http://www.factordb.com/aliquot.php?...&aq=4198862272 
20120403, 15:52  #8 
(loop (#_fork))
Feb 2006
Cambridge, England
6,353 Posts 
Have a play in gp: try running things of the form
Code:
for(u=10^12,10^12+10^5,s=u*2^3*3*5;if(gcd(s,2^4*3^2*5^2)==120,S1=sigma(s)s;S2=sigma(S1)S1;S3=sigma(S2)S2;S4=sigma(S3)S3;if(gcd(S4,15)==1,print(u,factor(u),factor(S1),factor(S2),factor(S3),factor(S4))))) You never lose the 2^3 on the first step. If you've lost it on the second step, you've gone via 2^3*3*5^W*X^2*p where W is 2 or more and X can either be 1 or something larger than seven. In 'actual escape' cases where you lose the 3 and 5, the trajectory seems always to be * get an extra power of five and have the remainder be a prime (2^3*3*5^2*p) * lose the extra power of five in exchange for changing the power of two (2^k*3*5) * have the power of five change * lose the three and this happens with probability about 0.0002 for numbers around 15 digits and 0.0001 for numbers around 25 digits. Of course you have higher probability (I think about 0.003) of escape routes that take a little longer, but the first step on an escape route appears always to be 2^3*3*5^2*p. (try working it through with 120*50221*389318203*51145795021) (or 120*1000000009991 where you lose the 2^3 only for a short period) 
20120403, 16:02  #9 
Jun 2003
1001000111011_{2} Posts 

20120403, 16:17  #10 
Apr 2010
Over the rainbow
2·3·5·79 Posts 
If my memory serve, it require that 2^3*3^2*5^2 and every other prime 1 mod 4.

20120403, 17:57  #11  
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
3·419 Posts 
Quote:
Sequence starting value = 3*2[sup]164[/sup] for merging with the following Aliquot sequence = 772032 Straight downdriver run quite directly from 95 digits to 7 digits in the size Stargate38, why this number 4198862272 = 2^{6}*11*5964293 what's special with this number? The number that I mentioned above, which is as well as, since it is being struck at C110, at the iteration number 3324 at all, this is again being a C110, I'd like to see it as well as being factored, again, thereby being gone ahead Code:
C110 = 22570951157462245129386607801622813637611303162132159250374656707973168765663980862177943354450043517261333717 at all yet still Last fiddled with by wblipp on 20120403 at 18:10 Reason: add code box 

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