20210808, 12:36  #1310  
"Alexander"
Nov 2008
The Alamo City
2^{3}×97 Posts 
Quote:
As an aside, I'd eventually like to reformat the conjecture statements to use proper superscripts and <var> tags. 

20210808, 14:22  #1311 
"Alexander"
Nov 2008
The Alamo City
2^{3}·97 Posts 
210^56 terminates at a prime. I'll let you know if I can get to any others before Monday.

20210808, 17:42  #1312  
"Garambois JeanLuc"
Oct 2011
France
2^{4}×43 Posts 
Quote:
I think I've corrected all that except for what you say : "You may want to consider using centered dots instead of periods." I don't understand what that means ? Quote:
I can't get into it right now... 

20210809, 00:26  #1313  
"Alexander"
Nov 2008
The Alamo City
2^{3}·97 Posts 
Quote:
That can actually be done using a regexcapable text editor (or sed) relatively quickly. I think I did most of the <var> tags already, so it would just leave the superscripts. 

20210809, 01:24  #1314 
"Alexander"
Nov 2008
The Alamo City
2^{3}·97 Posts 
I completed the trivial sequences for the two remaining Lehmer five bases up to 100 digits, so they'd be in by the Monday deadline.

20210809, 07:13  #1315 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
5940_{10} Posts 

20210809, 10:06  #1316 
"Garambois JeanLuc"
Oct 2011
France
1260_{8} Posts 
Thank you all for your recent (and older !) calculations.
I've recapped everything for tonight's big FactorDB scan that will start after 6 PM GMT. Here is the input table for my program that will be considered for this scan : Code:
basepuis=[[2,559],[3,335],[5,250],[6,210],[7,190],[10,160],[11,160],[12,150],[13,145],[14,140],[15,140],[17,140],[18,140],[19,140],[20,100],[21,100],[22,100],[23,100],[24,100],[26,100],[28,100],[29,100],[30,100],[31,100],[33,100],[34,100],[35,100],[37,100],[38,100],[39,100],[40,100],[41,100],[42,100],[43,100],[44,100],[45,100],[46,100],[47,100],[48,100],[50,100],[51,100],[52,100],[53,100],[54,100],[55,100],[58,100],[62,100],[72,100],[74,100],[75,100],[79,100],[98,100],[105,90],[162,80],[200,80],[210,80],[211,80],[220,80],[231,80],[242,70],[276,70],[284,70],[288,60],[338,60],[385,60],[392,60],[439,60],[450,60],[496,60],[552,60],[564,60],[578,60],[660,60],[720,60],[722,60],[770,60],[882,60],[966,60],[1155,50],[2310,50],[8128,40],[12496,40],[14288,40],[14316,40],[14264,40],[14536,40],[15472,40],[15015,40],[19116,40],[30030,40],[510510,30],[9699690,25],[33550336,25],[82589933,20],[223092870,20],[6469693230,20],[8589869056,20],[10^10+19,15],[200560490130,14],[7420738134810,14]] Of course, these 10 new bases will also be added on the project page, hopefully by the end of the week, if I have time ! Please check if the last bases you have calculated will be taken into account for the work this summer, and if no, let me know before the date and time specified above ! I will scan these 100 bases once. My program will also check for all sequences to see if any are broken. I hope I won't have such a bad surprise !!! 
20210809, 11:09  #1317  
"Garambois JeanLuc"
Oct 2011
France
2^{4}·43 Posts 
I just want to ask two questions about the demonstrations of the conjectures stated around this project.
1) I am totally unable to do it myself, but wouldn't it be possible to prove conjecture (10) by a method similar to the one used for conjecture (2) ? (see henryzz's post #476 and warachwe's post #480). Indeed, 2^121 = 3^2*5*7*13 and we do have the 13 which preserves the 7 for the second iteration, except perhaps for some values of k that I did not examine, but the whole problem is precisely there, isn't it ? 2) At the end of the quote below, Alexander says: "(the composite ones will be harder)". But does this mean that no one knows how to do such demonstrations yet, or does it just mean that we know how to do them, but that it is very long and tedious ? Quote:


20210809, 12:44  #1318 
Aug 2020
79*6581e4;3*2539e3
2^{2}×107 Posts 
I finished (2*18^2)^n = (648^2)^n up to n = 50.
For 50 < n < 60 the inial term as has 144169 digits and since every odd n has factor 3 they decrease very slowly. So I think I'll stop on that base now that it's at a nice n=50. 
20210809, 13:54  #1319  
"Alexander"
Nov 2008
The Alamo City
776_{10} Posts 
Quote:
Code:
306, 396, 696, 780, 828, 888, 996 Last fiddled with by Happy5214 on 20210809 at 13:54 

20210809, 16:13  #1320  
Aug 2020
19 Posts 
Quote:
This only happen when k is multiple of 13, for example 2^(12*13)1 =3^2*5*7*13^2*53*79*... This is why conjecture 34 ( 3^(18*37) ), 35 ( 3^(36*37) ), and 106 ( 11^(6*37) ) are false. But this doesn't mean all similar conjecture are false, as there maybe others prime(s) that preserve p. When we try to 'get rid of' those primes, there maybe yet another that will preserve p instead. Since the size of first iteration grow very quickly, it is hard to find other contradiction this way. If some of those are true, I imagine the proof might be similar to the proof of conjecture (2). 

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