20110726, 15:25  #1 
May 2011
France
7×23 Posts 
Beginning questions about Aliquot Sequences
What means 431060 411. sz 115 and what was the associate computetion.
John 
20110726, 16:36  #2  
"Ed Hall"
Dec 2009
Adirondack Mtns
2·3·619 Posts 
Quote:
* the sequence can also terminate in a cycle ** the index numbering starts at 0 

20110726, 17:13  #3  
Nov 2008
2×3^{3}×43 Posts 
Quote:


20110726, 18:39  #4 
May 2011
France
241_{8} Posts 
Always sorry
0 . 276 = 2^2 * 3 * 23
1 . 396 = 2^2 * 3^2 * 11 2 . 696 = 2^3 * 3 * 29 3 . 1104 = 2^4 * 3 * 23 4 . 1872 = 2^4 * 3^2 * 13 I don't understand how to compute 396 from 276 276 = 2^2 * 3 * 23 the sum of the factos is for me 4+3+23 =30 30 = 2*3*5 sum 10 10= 2*5 sum 7 7=7*1 end of the sequence so the sequands is 276,30,10,7,1 what I forget? John Last fiddled with by JohnFullspeed on 20110726 at 18:42 Reason: error on the values... 
20110726, 19:11  #5  
"Frank <^>"
Dec 2004
CDP Janesville
4112_{8} Posts 
Quote:
It might be easier to start with a smaller number. Say we're going to calculate the aliquot sequence for 12. If you plug 12 into Dairo's factorization applet, you get this answer: Quote:
So our sequence start out: 0. 12 = 2^2 * 3 Sum of divisors is 28, 2812 = 16 so the next line is: 1. 16 = 2^4 The sum of divisors is 1+2+4+8+16 = 32  16 (the original number) = 15. Continuing like this, our next couple of lines are: 2. 15 = 3 * 5 3. 9 = 3^2 4. 4 = 2^2 5. 3=3 And our sequence terminates. Does this help? 

20110727, 06:48  #6  
May 2011
France
7×23 Posts 
Aliquuot
Quote:
I need to study more but I sure thet your answer is rigth Now I can trry to code this algorithme tryiong to be the speeder Thanks John 

20110727, 08:57  #7 
May 2011
France
7×23 Posts 
Other Question
When the search is long:
searching to divisor or when the number of divisors(Nb ddigits)? Other? Tanks a Lot for your specification Perhaps you van add them in yout ' Get Sarted'. I thinnk that other strangers make the same mistake pime factor = divisor. Thanks Have you some chrono(the applet) Tosee the road it leaves to me..; John 
20110727, 09:13  #8  
"Frank <^>"
Dec 2004
CDP Janesville
2122_{10} Posts 
(I hope you don't mind, I moved these posts to their own thread....)
Quote:
There's a trick involved in searching when the numbers are large. Look up how the "sigma" function is calculated. (Hint: if you know the prime divisors of a number, the sum of the divisors is extremely easy to calculate....) Quote:
Quote:
http://www.alpertron.com.ar/ECM.HTM 

20110728, 05:59  #9 
May 2011
France
7·23 Posts 
Chrono??
I don't have a PC so i can't do it:
How many time to a search with 1 20 iterartioons 2 a value with 100 gigits 3 how many time to compute 400. 966 Thanks john 
20110728, 07:17  #10  
"Frank <^>"
Dec 2004
CDP Janesville
100001001010_{2} Posts 
Quote:
Quote:
Quote:
Quote:
[PS. Sorry, I had to put this one back on its primary task. The last 20 lines were going to take a while. Figure at least 12 hours at a minimum, getting longer as it approaches 400. So maybe ~12 hours to get the whole way.] Last fiddled with by schickel on 20110728 at 08:15 Reason: Adding PS 

20110728, 17:03  #11  
May 2011
France
7×23 Posts 
Aliquot
Quote:
TIPS : it is easy to compute the divisor sum sinc primes facto I use Sum = [ (p(a+1)  1) / (p  1) ] * [ (q(b+1)  1) / (q  1) ] * [ (r(c+1)  1) / (r  1) ] * ... You have better??? I go back when I will be faster: this afternoon or this evening 

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