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Old 2008-02-04, 19:29   #1
grandpascorpion
 
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Default Aliquot Sum Function

Hello,

Is there any known efficient heuristic or algorithm for finding positive numbers x such that aliquotSum(x) =n?

I have a similar one for the divisorSum but this is stumping me. divisorSum is much easier because you can just factor the n and work backwards.

Thanks.
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Old 2008-02-05, 12:31   #2
xilman
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Quote:
Originally Posted by grandpascorpion View Post
Hello,

Is there any known efficient heuristic or algorithm for finding positive numbers x such that aliquotSum(x) =n?

I have a similar one for the divisorSum but this is stumping me. divisorSum is much easier because you can just factor the n and work backwards.

Thanks.
Perhaps if you gave a (pointer to the) definition of the aliquotSum function you might get a better response.

Paul
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Old 2008-02-05, 15:01   #3
petrw1
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Based on a brief internet search I was led to believe that:
- aliquotsum = sum of all divisors of a number except it self (if aliquotsum(n) = n it is a perfect number)
- divisorsum = sum of ALL divisors including itself

If I read this right then the difference between aliquot(n) and divisorsum(n) is n.

Did you mean to say aliquotsum(x) = n rather than aliquotsum(n) = n ?
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Old 2008-02-05, 21:20   #4
Orgasmic Troll
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Quote:
Originally Posted by petrw1 View Post
Based on a brief internet search I was led to believe that:
- aliquotsum = sum of all divisors of a number except it self (if aliquotsum(n) = n it is a perfect number)
- divisorsum = sum of ALL divisors including itself

If I read this right then the difference between aliquot(n) and divisorsum(n) is n.

Did you mean to say aliquotsum(x) = n rather than aliquotsum(n) = n ?
I think he meant what he said.

Given n, he wants to find all x such that aliquotsum(x) = n

Equivalently, given n, find all x such that divisorsum(x) = n + x (which is a different question than "given n, find all x such that divisorsum(x) = n", which apparently is easy)

Last fiddled with by Orgasmic Troll on 2008-02-05 at 21:21
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Old 2008-02-05, 21:21   #5
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Exactly.

Last fiddled with by grandpascorpion on 2008-02-05 at 21:27
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Old 2008-02-06, 15:12   #6
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I worded this poorly earlier. The expression for the divisor sum (unless it's a perfect power) is always a reducible polynomial.

Can the same be said for the aliquot sum?
___________________________________________________________
Example:

Suppose n = a * b^2, where a and b are two different positive prime integers

The aliquot sum for n would be 1 + a + b + b^2 + a*b.

Is this irreducible?
_________________________________________________________

Thanks for any insight

Last fiddled with by grandpascorpion on 2008-02-06 at 15:16
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