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Old 2011-07-26, 15:25   #1
JohnFullspeed
 
May 2011
France

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Default Beginning questions about Aliquot Sequences

What means 431060 411. sz 115 and what was the associate computetion.
John
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Old 2011-07-26, 16:36   #2
EdH
 
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Quote:
Originally Posted by JohnFullspeed View Post
What means 431060 411. sz 115 and what was the associate computetion.
John
First, you must understand how an aliquot sequence is derived by adding the proper factors of each index and then factoring the new number, if possible. (If not possible, the sequence terminates.*) For example, look at the sequence for 431060 on the factordb site. You can note that the basic number (431060) is the starting point, the 411 is an index (in this case the last one), also referred to as a line or iteration** and the 115 describes the number of digits in the current index composite as opposed to the remaining composite, which in this case, presently, is 111 digits.

* the sequence can also terminate in a cycle
** the index numbering starts at 0
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Old 2011-07-26, 17:13   #3
10metreh
 
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Quote:
Originally Posted by JohnFullspeed View Post
What means 431060 411. sz 115 and what was the associate computetion.
John
Read the Getting started thread.
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Old 2011-07-26, 18:39   #4
JohnFullspeed
 
May 2011
France

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Default 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 2011-07-26 at 18:42 Reason: error on the values...
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Old 2011-07-26, 19:11   #5
schickel
 
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Quote:
Originally Posted by JohnFullspeed View Post
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
What's missing is that we use the divisors of the number, not just the prime divisors of the number.

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:
12 = 2 ^ 2 x 3

Number of divisors: 6

Sum of divisors: 28
As you can see, the prime divisors are 2 & 3, but it says there are 6 divisors. That's becuase the divisors are actually: 1, 2, 3, 4, 6, & 12. The sum is 28, but we subtract the number itself, since we want the aliquot divisors (aliquot divisor being defined as a number that divides the original number, excluding the number itself).

So our sequence start out:
0. 12 = 2^2 * 3

Sum of divisors is 28, 28-12 = 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?
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Old 2011-07-27, 06:48   #6
JohnFullspeed
 
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Default Aliquuot

Quote:
Does this help?
YES

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
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Old 2011-07-27, 08:57   #7
JohnFullspeed
 
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France

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Default 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
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Old 2011-07-27, 09:13   #8
schickel
 
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(I hope you don't mind, I moved these posts to their own thread....)
Quote:
Originally Posted by JohnFullspeed View Post
When the search is long:
searching to divisor or when the number of divisors(Nb ddigits)?
Other?
Do you mean how do you know when you're done?

There's a trick involved in searching when the numbers are large. Look up how the "sigma" \sigma(n) function is calculated. (Hint: if you know the prime divisors of a number, the sum of the divisors is extremely easy to calculate....)
Quote:
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.
That's why I moved these over to a new thread. Sometimes it can be the case that we're so at ease with what we do we don't know exactly what questions a newcomer might have.
Quote:
Thanks
Have you some chrono(the applet) Tosee the road it leaves to me..;
John
Ummm...not sure about this one--do you mean timings on factorizations? That depends on your PC. Or do you mean where is the applet? Sorry, didn't give the link before if that's the case:
http://www.alpertron.com.ar/ECM.HTM
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Old 2011-07-28, 05:59   #9
JohnFullspeed
 
May 2011
France

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Default 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
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Old 2011-07-28, 07:17   #10
schickel
 
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Quote:
Originally Posted by JohnFullspeed View Post
I don't have a PC so i can't do it:
How many time to a search with
1- 20 iterartioons
Depends on how big the number is. At the start of a sequence, too quick to measure. Higher up, could be months. I spent 6 weeks factoring a c157 for one of my sequence, so we could count that as 6 weeks for one iteration!
Quote:
2- a value with 100 gigits
Depending on how the number factors, it could be up to 4-5 hours, if the number splits 50-50.
Quote:
3- how many time to compute 400. 966
Do you mean 966 to 400 iterations? Using this system (2.6 GHz, Vista 64-bit, moderate load) it took about 35 minutes to get to iteration 385. I'll edit this later when it hits 100....
Quote:
Thanks
john
You're welcome!

[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 1-2 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 2011-07-28 at 08:15 Reason: Adding PS
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Old 2011-07-28, 17:03   #11
JohnFullspeed
 
May 2011
France

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Default Aliquot

Quote:
Do you mean 966 to 400 iterations? Using this system (2.6 GHz, Vista 64-bit, moderate load) it took about 35 minutes to get to iteration 385. I'll edit this later when it hits 100....
It's exactly what I want!!!!

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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