20170828, 15:09  #1 
"Garambois JeanLuc"
Oct 2011
France
2^{4}×43 Posts 
Factordb and aliquot sequences with useless size terms
I recently worked on aliquot sequences with record size terms.
The works are visible on this page, but sorry, in french : http://www.aliquotes.com/suite_terme...le_record.html Brief explanations of the principle before my two questions about factordb : It is quite easy to calculate several hundred terms of a strictly increasing aliquot sequence, each of those terms has more than 10 ^ 7 digits. If p is a prime and M=2^p1 is a prime too (a Mersenne prime), N = 2^(p1) * M, and N is a perfect number. So we have an aliquot sequence which starts on the number : j0 = N * z0 = 2^(p1) * M * z0. PGCD(M,z0)=1 because z0<M and M prime. So, j1 = σ(j0)  j0 = σ(N * z0)  N * z0 = σ(N) * σ(z0)  N * z0 = 2 * N * σ(z0)  N * z0 = N (2 σ(z0)  z0) = N * z1. As long as zi <M we have PGCD(M,zi)=1 We note that z1 = 2 σ (z0)  z0 and that, more generally, we will have z i+1 = 2 σ (zi)  zi. We note that z i+1> zi as long as zi<M. If N has millions of digits, then the terms of the aliquot sequence have millions of digits. I have 2 questions about factordb. 1) Question 1. It is possible to enter this aliquot sequence on factordb : 0 : 2^520 * (2^5211) * 3 = 2^520 * (2^5211) * 3^1 1 : 2^520 * (2^5211) * 5 = 2^520 * (2^5211) * 5^1 2 : 2^520 * (2^5211) * 7 = 2^520 * (2^5211) * 7^1 3 : 2^520 * (2^5211) * 9 = 2^520 * (2^5211) * 3^2 4 : 2^520 * (2^5211) * 17 = 2^520 * (2^5211) * 17^1 5 : 2^520 * (2^5211) * 19 = 2^520 * (2^5211) * 19^1 6 : 2^520 * (2^5211) * 21 = 2^520 * (2^5211) * 3^1 * 7^1 7 : 2^520 * (2^5211) * 43 = 2^520 * (2^5211) * 43^1 8 : 2^520 * (2^5211) * 45 = 2^520 * (2^5211) * 3^2 * 5^1 9 : 2^520 * (2^5211) * 111 = 2^520 * (2^5211) * 3^1 * 37^1 10 : 2^520 * (2^5211) * 193 = 2^520 * (2^5211) * 193^1 But it is not possible to enter this aliquot sequence on factordb : 0 : 2^74207281 * (2^742072811) * 3 = 2^74207281 * (2^742072811) * 3^1 1 : 2^74207281 * (2^742072811) * 5 = 2^74207281 * (2^742072811) * 5^1 2 : 2^74207281 * (2^742072811) * 7 = 2^74207281 * (2^742072811) * 7^1 3 : 2^74207281 * (2^742072811) * 9 = 2^74207281 * (2^742072811) * 3^2 4 : 2^74207281 * (2^742072811) * 17 = 2^74207281 * (2^742072811) * 17^1 5 : 2^74207281 * (2^742072811) * 19 = 2^74207281 * (2^742072811) * 19^1 6 : 2^74207281 * (2^742072811) * 21 = 2^74207281 * (2^742072811) * 3^1 * 7^1 7 : 2^74207281 * (2^742072811) * 43 = 2^74207281 * (2^742072811) * 43^1 8 : 2^74207281 * (2^742072811) * 45 = 2^74207281 * (2^742072811) * 3^2 * 5^1 9 : 2^74207281 * (2^742072811) * 111 = 2^74207281 * (2^742072811) * 3^1 * 37^1 10 : 2^74207281 * (2^742072811) * 193 = 2^74207281 * (2^742072811) * 193^1 I suppose the numbers are too big for factordb and factordb does not seem to know the prime number of mersenne (2^742072811), that's right ? Aliquot sequences with more than 700 terms available here : http://www.aliquotes.com/parfait_521_3.txt And with terms with more than 44 millions digits (!!!) http://www.aliquotes.com/parfait_74207281_3.txt And on factordb, the limit (130100 digits) : http://factordb.com/sequences.php?se...20&fr=0&to=100 2) Question 2. If we look at the 3 previous aliquot sequences (2^520 * (2^5211) * 3 and 2^74207280 * (2^742072811) * 3 and 2^216090 * (2^2160911) * 3), we note that the sequence of the numbers zi are identical : only the perfect number N changes. If I enter the first aliquot sequence in factordb, (see here : http://factordb.com/sequences.php?se...20&fr=0&to=100), the database "knows" all the prime numbers of the decompositions of the cofactors zi. Then why if I enter another starting number in factordb, like 2 ^ 606 * (2 ^ 6071) * 3, the database does not indicate the terms of the aliquot sequence whith the same cofactors zi ? 
20170828, 17:43  #2 
"Vincent"
Apr 2010
Over the rainbow
2^{2}×11×61 Posts 
The db has no idea of what a perfect number is, or what to do with it
Last fiddled with by firejuggler on 20170828 at 17:44 
20170829, 07:16  #3  
"Garambois JeanLuc"
Oct 2011
France
688_{10} Posts 
Quote:
I assumed that it would be possible for db "to know" the Mersenne prime numbers in their "condensed" form. For example, in the form : 2 ^ 74207281  1 Instead of that form which is written with more than 44 million digits : 451129962706 ... 557930315776 And especially that db is not obliged to check again the primality of this huge number first already known and famous. But it must be very difficult to program things like that. 

20170829, 07:28  #4 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3×29×83 Posts 
The factordb doesn't know anything about the theory behind aliquot sequences, and in particular doesn't know anything about how perfect numbers with "small" cofactors all have the same cofactor sequence. All it knows is that there's a whole bunch of big numbers, and it doesn't try to factor or otherwise test the big numbers.

20170830, 08:04  #5 
"Garambois JeanLuc"
Oct 2011
France
2^{4}×43 Posts 
OK, thank you for your answers !

20170830, 08:11  #6 
(loop (#_fork))
Feb 2006
Cambridge, England
2^{2}×3^{2}×179 Posts 
Given that the cofactors are the same whatever perfect number you're looking at (at least until you start getting the Mersenne prime as an accidental cofactor), why not compute the series of cofactors explicitly and just label the perfect number something symbolic?

20170830, 11:03  #7 
"Garambois JeanLuc"
Oct 2011
France
2^{4}·43 Posts 
This is exactly what I did to get for example this aliquot sequence whose terms have more than 44677235 digits in base 10 :
http://www.aliquotes.com/parfait_74207281_3.txt My problem is that I can not inform the database factordb with this aliquot sequence because factordb does not know the "symbolic" representation of the large prime numbers of Mersenne. 
20170830, 11:47  #8  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:


20170831, 19:03  #9  
"Garambois JeanLuc"
Oct 2011
France
2^{4}×43 Posts 
Quote:
OK, thank you ! You are right, I tried with success for the 35th Mersenne prime. But no success for the 36th Mersenne prime ! I didn't know. 2^1398268*(2^13982691)*3 : SUCCESS 2^2976220*(2^29762211) * 3 : NO SUCCESS 

20170901, 00:06  #10  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·4,813 Posts 
Quote:
By uploading uselessly large algebraically trivial results into factordb, you are sabotaging users who use factordb for proper reasons. 

20170902, 00:21  #11 
"Garambois JeanLuc"
Oct 2011
France
1260_{8} Posts 

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