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?
Amicable Pair $(a, b)$: given $a$, what are limits on size of $b$?
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number-theory
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0m 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? – 2010-11-13
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0One 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 $n$... – 2012-03-05
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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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0The 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. – 2010-11-13
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0What is that exponent of e? – 2010-11-14
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0@NotSuper It's the Euler-Mascheroni constant: http://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant – 2010-11-14