20151128, 16:31  #67 
"Robert Gerbicz"
Oct 2005
Hungary
577_{16} Posts 
My method to obtain more than 7 million new amicable pairs was pretty standard:
First generate some (i,1) type of amicable pairs, then use the easy (unpublished) te Riele rule get much more pairs. The search was especially successful on (4,1) type, found more than 900 in this type. (previously it was known only 600 such pairs). A gold mine was: 41 Gerbicz 2015 4042515853710748097920965379860021278894537700264375=3^2*5^4*13*43*193*1499*9821213*35934768372149*12591529154973463187 4045213074037534292614128569338790912846756018135625=3^2*5^4*13*43*193*6665761314523585756066383974737009258199999 it has given a record of 720344 new amicable pairs. 
20151129, 17:13  #68  
Jun 2015
Stockholm, Sweden
83 Posts 
Quote:
Last fiddled with by Sergei Chernykh on 20151129 at 17:15 

20151130, 08:32  #69  
"Robert Gerbicz"
Oct 2005
Hungary
577_{16} Posts 
Quote:


20151201, 07:31  #70  
Mar 2015
Australia
2×41 Posts 
Quote:
Also can you say from your work or others that all of the regular (i,1) pairs have been found up to a particular size? (Obviously to 17 digits). Andrew PS Is the "unpublished" rule in one of Garcia's papers? Last fiddled with by AndrewWalker on 20151201 at 07:33 

20151201, 07:36  #71  
Jun 2015
Stockholm, Sweden
83 Posts 
Quote:
41 Chernykh 2015 144296691213180636675=3^3*5^2*19*43*113*743*773*1129*3571 144846519073526051325=3^3*5^2*19*43*113*2324362124159 It looks like that all (i, 1) pairs up to 20 digits are already found. All currently known (i, 1) 20 digit pairs were found before 2007. 

20151201, 11:51  #72  
"Robert Gerbicz"
Oct 2005
Hungary
1399_{10} Posts 
Quote:
I have searched "only" for regular (i,1) types as these gives much more pairs than the irregulars (note that you can apply the te Riele rule also for the irregulars). Quote:
say the 900 regular (4,1) pair has given 7M new pair (don't know the correct figure), in average that is less than 8000 new pairs per one regular (i,1) pair. 

20151211, 19:01  #73 
"Robert Gerbicz"
Oct 2005
Hungary
1,399 Posts 
So far tripled the known number of amicable pairs in this year!
Finally downloaded all c2 files, it took some time, the total size is 25.5 GBytes (the real count is obviously somewhat larger). Used curl.h library to download it with a c code. http://sech.me/ap/dl.php?d=11&a=55&b=59 is also downloading the 11 digits pairs starting with 54 in the smaller term, what is a problem (not checked every a,b combinations). And this is an error not only for 11 digits. My suggestion to make the upload (and optionally the download) faster on http://sech.me/ap/submit.html, while keeping the c2 file and upload format. It would be even an improvement on David Moews's format, see that on http://djm.cc/aliquotdatabase/aliquotdatabase.uhtml . Let (n,m) be an amicable pair (n<m) and g=gcd(n,m), in the upload: in the first line there is no change ([optional #] type author year) in the second line give the prime factorization of g. in the third line give the prime factorization of n/g and m/g, seperated by a single space. the fourth line is empty. Let p,q be the largest prime factors of n and m, in the general case in the prime factorization these has got an exponent=1, and in this case we can skip these factors in the third line, as we can recover them quickly: we know a,b where n=a*p m=b*q here gcd(a,p)=gcd(b,q)=1, as p,q are prime factors with exponent=1. Since n,m is an amicable pair: sigma(a)*(p+1)=sigma(b)*(q+1)=a*p+b*q from this (linear equation system) we can easily get that: p=(b*sigma(b)+sigma(a)*(sigma(b)b))/(a*sigma(b)+sigma(a)*(bsigma(b))) q=sigma(a)*(p+1)/sigma(b)1 Note: if the denominator=0 in the formula for p, then this gives no solution, otherwise the numerator should be also equal to zero (so before division it was c1*p=c0, where c1=c0=0) , but then (a+b)*sigma(b)=0, contradiction. So you should consider only the case where denominator!=0. One example: instead of the long form: 52 Chernykh 2015 10000038430333472389639196495581571703811508519341139357606404362100114924718480425327302956481552091613122511439512045864644400554493484575018567812942028023870853758600890282709934135211420149375925=3^3*5^2*19*31*647*2459*5498256668803134744400284381645697*2159177*15595178728169*80102078917417*142954673388881503354407393922788121*7457169855869184777027175733414646695449141582155087600336262807138347680833561875599 10000043061746313675688063566993046183019639196803558251029615447935398543993674990250570277297741997696844921859194700501549755659015120076467874934408892947993132186621683421380108987795019274624075=3^3*5^2*19*31*647*2459*5498256668803134744400284381645697*385585725761869156211745483928670847042430834744924574107532526867599*7457169855869184005855902789575548927338966166838568239579444556765289145410452559759 submit it as: 52 Chernykh 2015 3^3*5^2*19*31*647*2459*5498256668803134744400284381645697 2159177*15595178728169*80102078917417*142954673388881503354407393922788121 385585725761869156211745483928670847042430834744924574107532526867599 There are very few amicable pairs where p or q has got an exponent>1, for example: n=46129121115;m=54412443813; for these pairs just use the full factorization of both(!) number on the third line. So for example write: X44 Moews&Moews 1992 3*7^2*31 5*11*41*67^2 13*17*97*557 This also means that in every case you should check these two forms, and in general the full factorization form does not give a further solution as for example sigma(n)=n+m won't be true. I would be also interested in a zipped download that is using the above stripped format. Say make it only once a week. Any idea/suggestion? ps. as still more than 90% of all pairs is in te Riele rule, we could use this. But it would be a harder job. And for me now the above stripped format is enough. 
20151212, 16:47  #74  
Jun 2015
Stockholm, Sweden
83 Posts 
Quote:
Quote:
P.S. I can set up weekly cron job on the server which saves the database in this form. Last fiddled with by Sergei Chernykh on 20151212 at 17:18 

20151212, 19:34  #75  
"Robert Gerbicz"
Oct 2005
Hungary
1,399 Posts 
Quote:
Just to correct my above post: in the general case let p and q be the largest prime factor of n/g and m/g. And with your idea: just select any p,q prime factor (with exponent=1) of n/g and m/g in special cases. What I have not written above that if n/g or m/g is greater than 1 and it is a powerful number (https://en.wikipedia.org/wiki/Powerful_number) then trivially you can't select such a prime factor. Not checked the database, but even if there is no such known amicable pair it could exist. And for a coprime amicable pair print 1 in the second line (g=gcd(n,m)=1), we still don't know if this exist. 

20151215, 18:18  #76  
Jun 2015
Stockholm, Sweden
83 Posts 
Quote:
dl.php is still available though, if you use it for automatic download scripts. P.S. If you don't care about the order of pairs in files, you can download raw text files, i.e. sech.me/ap/100/23.txt for all 100digit pairs starting with 23. Last fiddled with by Sergei Chernykh on 20151215 at 18:21 

20151217, 14:34  #77 
"Robert Gerbicz"
Oct 2005
Hungary
1,399 Posts 

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