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 2008-02-04, 19:29 #1 grandpascorpion     Jan 2005 Transdniestr 503 Posts 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.
2008-02-05, 12:31   #2
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Quote:
 Originally Posted by grandpascorpion 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

 2008-02-05, 15:01 #3 petrw1 1976 Toyota Corona years forever!     "Wayne" Nov 2006 Saskatchewan, Canada 13·349 Posts 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 ?
2008-02-05, 21:20   #4
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Quote:
 Originally Posted by petrw1 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

 2008-02-05, 21:21 #5 grandpascorpion     Jan 2005 Transdniestr 503 Posts Exactly. Last fiddled with by grandpascorpion on 2008-02-05 at 21:27
 2008-02-06, 15:12 #6 grandpascorpion     Jan 2005 Transdniestr 503 Posts 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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