Beginning questions about Aliquot Sequences
 

What means 431060 411. sz 115 and what was the associate computetion.
John

[QUOTE=JohnFullspeed;267581]What means 431060 411. sz 115 and what was the associate computetion.
John[/QUOTE]
First, you must understand how an [URL="http://en.wikipedia.org/wiki/Aliquot_sequence"]aliquot sequence[/URL] 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 [URL="http://www.factordb.com/sequences.php?se=1&eff=2&aq=431060&action=all&fr=0&to=100"]431060 on the factordb site[/URL]. 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.

[SIZE=1]* the sequence can also terminate in a cycle
** the index numbering starts at 0[/SIZE]

[QUOTE=JohnFullspeed;267581]What means 431060 411. sz 115 and what was the associate computetion.
John[/QUOTE]

Read the [url=http://www.mersenneforum.org/showthread.php?t=12140]Getting started[/url] thread.

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

[QUOTE=JohnFullspeed;267600]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[/QUOTE]What's missing is that we use the [b]divisors[/b] of the number, not just the [b][i]prime[/i] divisors[/b] 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[/quote]As you can see, the prime divisors are 2 & 3, but it says there are [b]6[/b] 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, [i]excluding the number itself[/i]).

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 ([i]the original number[/i]) = 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?

Aliquuot
 

[QUOTE]Does this help? [/QUOTE]

YES

I need to study more but I sure thet your answer is rigth:smile::smile::smile::smile::smile:
Now I can trry to code this algorithme tryiong to be the speeder:smile::smile::smile:

Thanks
John

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

(I hope you don't mind, I moved these posts to their own thread....)[QUOTE=JohnFullspeed;267667]When the search is long:
searching to divisor or when the number of divisors(Nb ddigits)?
Other?[/quote]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" [tex]\sigma(n)[/tex] function is calculated. (Hint: [spoiler]if you know the prime divisors of a number, the sum of the divisors is extremely easy to calculate....[/spoiler])[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.[/quote]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[/QUOTE]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:
[url]http://www.alpertron.com.ar/ECM.HTM[/url]

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

[QUOTE=JohnFullspeed;267736]I don't have a PC so i can't do it:
How many time to a search with
1- 20 iterartioons[/quote]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[/quote]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[/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....
[quote]
Thanks
john[/QUOTE]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.]

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....[/QUOTE]

It's exactly what I want!!!!

TIPS : it is easy to compute the divisor sum sinc primes facto

I use
[I]Sum[/I] = [ ([I]p(a+1)[/I] - 1) / ([I]p[/I] - 1) ] * [ ([I]q(b+1)[/I] - 1) / ([I]q[/I] - 1) ] * [ ([I]r(c+1)[/I] - 1) / ([I]r[/I] - 1) ] * ...

You have better???

I go back when I will be faster: this afternoon or this evening:smile::smile::smile::smile: