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Let any Amicable Pair be represented by $(a, b)$, with $a \lt b$. In a search for Amicable Pairs, let $(n, m)$ with $n < m$, be a candidate to be tested for being an amicable pair. Given $n$, is there way to calculate both the least upper bound and greatest lower bound for the m that would make $(n, m)$ an Amicable Pair?

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m has to be equal to the sum of the proper divisors of n. Or do you mean you are given the approximate size of n? –  Qiaochu Yuan Nov 13 '10 at 11:34
    
One thing worth noting: since as Derek notes, $(m,n)$ is an amicable pair only if $m=\sigma(n)-n$ (and vice versa), there really isn't a lot of 'searching' to be done; there's only one value of $m$ that needs to be tested for each value of $n$... –  Steven Stadnicki Mar 5 '12 at 4:54

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An Amicable pair of numbers $(m,n)$ is such that each is equal to the sum of the proper divisors of the other, so if you know $n$ then $m=\sigma(n)-n,$ where $\sigma(n)$ is the usual sum of divisors function. Hence you are essentially asking for bounds on $\sigma(n).$

Now $\sigma(n)-n=1$ when $n$ is prime and so for a lower bound $m \ge 1,$ since $\sigma(n) \ge n+1$ for all $n \ge 2.$

The following upper bound for $\sigma(n)$

$$\sigma(n) < e^\gamma n \log \log n + \frac{ 0.6482 n}{\log \log n}$$

was given on mathoverflow.

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The upper bound mentioned should of course have the condition $n \ge 3$ but I'm not risking an edit as some nightmare changes with regard to parsing seem to have taken place on this site recently. –  Derek Jennings Nov 13 '10 at 13:33
    
What is that exponent of e? –  NotSuper Nov 14 '10 at 3:02
    
@NotSuper It's the Euler-Mascheroni constant: en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant –  Derek Jennings Nov 14 '10 at 8:45
    
Thanks! –  NotSuper Nov 15 '10 at 6:26

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