20120513, 23:45  #1948 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3×29×83 Posts 
C112?
Seems no one's doing it. I'll do it, shouldn't take more than an hour or two.

20120514, 02:51  #1949 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3×29×83 Posts 
Close Lines
Line 4824
Line 4825 Unfortunately, "bc  An arbitrary precision calculator language" returned 0 for (largesmall)/small, and I have not the slightest clue to pari/gp. (Anybody who cares can surely figure it out on their own.) Edit: 4826 is pretty close as well, though not quite as close as the first two. Last fiddled with by Dubslow on 20120514 at 03:38 
20120514, 04:20  #1950 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
1C35_{16} Posts 
C122
I'll have NFS on the C122 (line 4826) done in 1213 hours.
Last fiddled with by Dubslow on 20120514 at 04:21 Reason: () 
20120514, 05:20  #1951  
"Frank <^>"
Dec 2004
CDP Janesville
2×1,061 Posts 
Quote:
Clifford records the highest escape from 2^4 * 31 as 129 digits in 5778. I escaped it at 113 digits in 48462 via this factorization: Code:
1236 . c113 = 2^4 * 31^2 * 6188785238747719 * 230402350198068564832070130054678954206970878230960200048690011877198238894335419716602085901 1237 . c113 = 2^5 * 31 * 71 * 746231 * 144030777132837510217987539096343417010500889513 * 2902096090182134830390738125010054022390760487434126181 PS. For those of you that haven't seen it yet, here's the graph (2 troughs at 33 and 35 digits): 

20120514, 06:22  #1952  
Romulan Interpreter
Jun 2011
Thailand
5^{2}×7×53 Posts 
Quote:
Or should I pray for 618480 (1856: C140=2^4*3^3*...*C134) to get a 31?? (and then pray harder to lose it in few terms? hehe) Last fiddled with by LaurV on 20120514 at 06:24 

20120514, 06:41  #1953  
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
16065_{8} Posts 
Quote:
Quote:
Last fiddled with by Dubslow on 20120514 at 07:13 Reason: gotta love the smileys 

20120514, 09:16  #1954 
"Robert Gerbicz"
Oct 2005
Hungary
5×17^{2} Posts 
Escaping from 2^4*31 isn't very hard, see the elf file for 5778. Between line 390973 we have 2^4*31^e, for e>0. And for 21 times e=2, for one time e=3 and for the rest e=1. So escaping it is easy but on the next line we see again 2^4*31. I would say it is hard to kill the 31 as a factor in 2^4*31.

20120514, 09:30  #1955  
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3×29×83 Posts 
Double Post
Quote:
Edit: I have the last post, so I'll use it. That C122 split as a P59*P63, a pretty nice split. In the meantime, the C124 is almost ready for NFS; Code:
sum: have completed work to t38.16 Last fiddled with by Dubslow on 20120514 at 09:49 Reason: Adding post 2 

20120514, 16:33  #1956 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3^{2}×17×61 Posts 
From the tail of sequence 4788, it appears that 31 gets raised to the square only in 1/(2*31) iterations. (Not 1/31 that one would expect from a random process.)
From the same tail, the average gain in every 62 iterations is 10.7 digits (1.49x per iteration). 
20120515, 10:14  #1957 
"Tapio Rajala"
Feb 2010
Finland
3^{2}·5·7 Posts 
I advanced the sequence by a couple of iterations. I will not run NFS on the current c124 @ i4841, if it comes to that (so far I have run ECM 800@1M + 300@3M).

20120610, 15:08  #1958 
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
3×419 Posts 
What happened to the aliquot.de page? It has not been updated since the April 1 day at all?
Does not contain information regarding the advance for the 4788 (aka 314718) sequence more than 2000 lines; nor the merge of the 345324 sequence (maybe this was being the last one as of now? 9230 open end sequences being remaining below 1000000?) Mr. Creyaufmüller been on over a long vacation, or otherwise being suffering from a major illness, sorry? I see that Mr. Clavier's page as well as has got no information relative to the advance for this 4788 sequence, nor since the 9282 sequence downdriver capture event at all, but this is not being worth mentioning at all; since this is being relatively a new incident. Can the condition be given to escape (to break away) off from the 2^{4}*31 driver, to be needed as such? (31 being raised to the higher even powers is being very rarer enough, though) Iteration line factoring into The next subsequent line mutates into 2^{4}*31^{2}*(1 mod 4 prime) 2*31 2^{4}*31^{2}*(3 mod 8 prime) 2^{2}*31 2^{4}*31^{2}*(7 mod 16 prime) 2^{3}*31 2^{4}*31^{2}*(15 mod 32 prime) 2^{>4}*31 2^{4}*31^{2}*(1 mod 4 prime)*(1 mod 4 prime) 2^{2}*31 2^{4}*31^{2}*(1 mod 4 prime)*(3 mod 8 prime) 2^{3}*31 2^{4}*31^{2}*(1 mod 4 prime)*(7 mod 16 prime) 2^{>4}*31 2^{4}*31^{2}*(3 mod 8 prime)*(3 mod 8 prime) 2^{>4}*31 2^{4}*31^{2}*(1 mod 4 prime)*(1 mod 4 prime)*(1 mod 4 prime) 2^{3}*31 2^{4}*31^{2}*(1 mod 4 prime)*(1 mod 4 prime)*(3 mod 8 prime) 2^{>4}*31 2^{4}*31^{2}*(1 mod 4 prime)*(1 mod 4 prime)*(1 mod 4 prime)*(1 mod 4 prime) 2^{>4}*31 Even the perfect square factoring or otherwise twice perfect square line is being very highly rare enough that they mutate into an odd number, leading furthermore into very rapid terminating finishes off, away Last fiddled with by Raman on 20120610 at 16:08 
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