20080204, 19:29  #1 
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. 
20080205, 12:31  #2  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2·3·1,753 Posts 
Quote:
Paul 

20080205, 15:01  #3 
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 ? 
20080205, 21:20  #4  
Cranksta Rap Ayatollah
Jul 2003
1010000001_{2} Posts 
Quote:
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 20080205 at 21:21 

20080205, 21:21  #5 
Jan 2005
Transdniestr
503 Posts 
Exactly.
Last fiddled with by grandpascorpion on 20080205 at 21:27 
20080206, 15:12  #6 
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 20080206 at 15:16 
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