20160930, 07:43  #34 
"Andrew Booker"
Mar 2013
125_{8} Posts 

20161001, 18:21  #35 
Oct 2011
484_{10} Posts 

20161002, 23:19  #36  
"Andrew Booker"
Mar 2013
85_{10} Posts 
Quote:
I've extended my earlier computations to include all sociable cycles below 10^{18} and all amicable pairs below 10^{14}, except for one that overflowed the 2^{63} limit. Some interesting graphs emerged, such as the attached one with 58 nodes. Last fiddled with by arbooker on 20161002 at 23:59 

20161003, 16:27  #37 
Oct 2011
2^{2}·11^{2} Posts 
Arbooker,
I am trying to calculate the numbers of aliquots antecedents for all even numbers between 10^11 and 10^11+10^7. It will take me 4 months. If I understand, with your algorithm, you could do this in a few days ? Your work really interest me. I would like to see your algorithm ! 
20161003, 17:34  #38 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2·2,909 Posts 
It should be possible to compress this sort of data quite well I imagine. Otherwise it would take petabytes. Maybe it isn't possible to get it down to nearer a terabyte.
How common is it for a 14 digit number to be part of a finite isolated subgraph? Do most numbers balloon into having a subgraph which is too large to calculate? To me this sort of thing could be as interesting as cycles. Last fiddled with by henryzz on 20161003 at 17:34 
20161004, 23:45  #39 
"Andrew Booker"
Mar 2013
5·17 Posts 
Perhaps, but for working out the antecedents of so many numbers I think it would be better to compute s(n) for all possible n. With a windowed sieve you can factor all the numbers up to 2(10^{11}+10^{7}) in a matter of hours. I can give you some generic code for this if it would help.

20161005, 14:00  #40  
Oct 2011
2^{2}×11^{2} Posts 
Quote:
I want to see if it is faster than mine... 

20161021, 12:00  #41  
"Andrew Booker"
Mar 2013
5·17 Posts 
Quote:


20161021, 13:36  #42  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
Last fiddled with by science_man_88 on 20161021 at 14:23 

20161021, 20:26  #43  
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2·2,909 Posts 
Quote:
However, I don't understand how the probability of the kth iterate of s^1 containing an odd number >> 1/k. My first thought is that the probability of the 2nd iterate is something like the probability of the 1st iterate squared. Maybe I am on the wrong track. I suppose that this doesn't take into account that if the first iterate has an odd number the second iterate won't be empty and will definitely have an odd number. I need to think on this further when it is not evening. It does now seem to be laid out a bit better than yesterday. The figures/algorithms aren't in the middle of paragraphs anymore. 

20161022, 08:14  #44 
"Andrew Booker"
Mar 2013
5·17 Posts 
Thanks for your careful reading. What I really mean there is that the chance of getting a p+1 is >> 1/k. That's based simply on the fact that the numbers grow at most like 2^{k}n, and primes occur with density 1/log.
Last fiddled with by arbooker on 20161022 at 08:15 
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