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Old 2011-07-28, 18:00   #12
10metreh
 
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Quote:
Originally Posted by JohnFullspeed View Post
I use
Sum = [ (p(a+1) - 1) / (p - 1) ] * [ (q(b+1) - 1) / (q - 1) ] * [ (r(c+1) - 1) / (r - 1) ] * ...
As long as by p(a+1) you mean pa+1, I'm pretty sure that's the one schickel was thinking of.
Now for a simple exercise: can you prove it? (Dr Silverman would probably say it was trivial )
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Old 2011-07-28, 22:24   #13
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Quote:
Originally Posted by schickel View Post
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: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?
this looks nothing like what I thought ! I know how to code what I know in PARI.
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Old 2011-07-29, 00:17   #14
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Quote:
Originally Posted by JohnFullspeed View Post
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???
http://fr.wikipedia.org/wiki/Suite_aliquote , à tout hasard.
Veuillez aller au "Relation de récurrence" et voilà!
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Old 2011-07-29, 06:17   #15
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For information it's the ssame with recurence

Thanks to search;
John
(It's the morning in France : I attack 966 from 0 to 122 :last 64 bits (19 digits))
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Old 2011-07-29, 13:55   #16
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Quote:
Originally Posted by JohnFullspeed View Post
For information it's the ssame with recurence

Thanks to search;
John
(It's the morning in France : I attack 966 from 0 to 122 :last 64 bits (19 digits))
You are doing this for your edification, correct?

(You do realize that all the sequences with a starting number <1.000.000 have been worked to >100 digits, right?)
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Old 2011-07-29, 18:44   #17
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In fact I write personal tools: list of primes, factoring, primarily test

I don't want a generic code l my computer is a 64 bits so less 10^20 yous use the ROM after I write the add, div, mod,sort,sqrt...
I extract a sqr of a number of 1 000 000 digits
Just for the fun
A modulo for 1500 digits is not useful but funny if it needs 2 or 3 seconds

In fract Iuse 966 to verify py code. I have a little pb with the itrtations 70....


To be quiet I try value> 1.000.000 Is there better values to begin an Aliquot
sequence?

It's now evening My Aliquot search is write. Tomorrow I attack more 20 digits
John
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Old 2011-07-30, 13:05   #18
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Quote:
Originally Posted by JohnFullspeed View Post
In fact I write personal tools: list of primes, factoring, primarily test

I don't want a generic code l my computer is a 64 bits so less 10^20 yous use the ROM after I write the add, div, mod,sort,sqrt...
I extract a sqr of a number of 1 000 000 digits
Just for the fun
A modulo for 1500 digits is not useful but funny if it needs 2 or 3 seconds

In fract Iuse 966 to verify py code. I have a little pb with the itrtations 70....


To be quiet I try value> 1.000.000 Is there better values to begin an Aliquot
sequence?

It's now evening My Aliquot search is write. Tomorrow I attack more 20 digits
John
well :
Code:
trying(x,y)=b=1;a=[x];until(b==y || isprime(x)||sigma(x)-x==x,if(isprime(x),break(),x=sigma(x)-x;a=concat(a,x);b=b+1));a
is my best so far even with the counter variable. in fact I have another script that does basically the same thing but this is about 30 + times faster most times.
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Old 2011-07-30, 14:31   #19
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Quote:
Originally Posted by JohnFullspeed View Post
In fact I write personal tools: list of primes, factoring, primarily test

I don't want a generic code l my computer is a 64 bits so less 10^20 yous use the ROM after I write the add, div, mod,sort,sqrt...
I extract a sqr of a number of 1 000 000 digits
Just for the fun
A modulo for 1500 digits is not useful but funny if it needs 2 or 3 seconds

In fract Iuse 966 to verify py code. I have a little pb with the itrtations 70....


To be quiet I try value> 1.000.000 Is there better values to begin an Aliquot
sequence?

It's now evening My Aliquot search is write. Tomorrow I attack more 20 digits
John
Actually, to test and verify, why not use another interesting one? Try 3630 and see what happens.

As far as start values over 1.000.000, there is no organized project there, but we can suggest many values under that that would be owrth pursuing.
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Old 2011-07-30, 15:13   #20
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Quote:
Originally Posted by science_man_88 View Post
well :
Code:
trying(x,y)=b=1;a=[x];until(b==y || isprime(x)||sigma(x)-x==x,if(isprime(x),break(),x=sigma(x)-x;a=concat(a,x);b=b+1));a
is my best so far even with the counter variable. in fact I have another script that does basically the same thing but this is about 30 + times faster most times.
There isn't any point in using PARI for practical aliquot sequence computations as its factorization is much slower than, say, Yafu. Programming the calculation of the next term isn't the difficult bit.
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Old 2011-07-30, 17:53   #21
JohnFullspeed
 
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Quote:
Originally Posted by schickel View Post
Actually, to test and verify, why not use another interesting one? Try 3630 and see what happens.
Because I have all the iterations!
I can see that I have a problem with iteration 65:
i have the good sum but I find a bad factoring;

I try 3630.....perdect,... or....
John
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Old 2011-07-31, 05:18   #22
JohnFullspeed
 
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Quote:
Originally Posted by science_man_88; [CODE

trying(x,y)=b=1;
a=[x];
until(b==y ||isprime(x)||sigma(x)-x==x,
if (isprime(x),break(),x=sigma(x)-x;
a=concat(a,x);
b=b+1))
;a
[/CODE]
in fact I have another script that does basically the same thing but this is about 30 + times faster most times.
I don't understand the algorithm can you explain me
and if you have a faster I'm interested!!!!!

where :Dairo's factorization applet,


John

Last fiddled with by JohnFullspeed on 2011-07-31 at 05:25
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