Let $(n, m)$, $n < m$ be an Amicable Pair. Looking for large sets of integers that cannot be $n$. I've got a Python script that grinds out Amicable Pairs by trying out each integer: $i, i+1, i+2,\ldots$ . I've now found all APs with $1 \leq n \leq 65,000,000$. At this point I'm finding that it takes $24$ minutes per each additional million. To get into much larger $n$ I'll need to find a way to filter out substantial numbers of $n$ in advance. Have any been discovered?   
My list of the first 202 APs: http://tutoree7.pastebin.com/dpEc0RbZ
 A: Starting off by giving advice that does not seem to directly answer your question, you
could do worse than to look at a few of the papers that bound the
number of amicable numbers, particularly Erdos' 1955
paper showing that the
density of amicable number is 0.  This paper, is simple, direct, and
largely the inspiration of what follows.  Also worth looking at are
Carl Pomerance's
pair of
papers giving
better bounds.  These papers are considerably more difficult than the Erdos paper, and at
least for me, less applicable to inspiring good code.  There is also a
pair of papers by Erdos on $\sigma(n)/n$ that I found particularly
illuminating, but I cannot find them (my wife may have a point about
the state of my office.)
More directly relevant is H. J. J. te Riele's
paper
on the search for amicable numbers less than $10^{10}$
Now to more practical algorithmic suggestions.  The ideas that follow, along
with the details for testing that $\sigma(n)=m$ sketched
here
are responsible for most of the known amicable
pairs between $10^{11}$ and
$2^{64}$.
First, you only want to search for the smaller member of each amicable
pair, this allows you to ignore numbers you know will not be abundant,
and as around 3/4 of all numbers are not abundant this can speed
things dramatically.
The main idea is to forget that the integers are well ordered
(pinheaded enumerationist propaganda) and search through the integers
in the range you are interested in by building them up by their prime
factorization, essentially a tree traversal.  This has many advantages, computing $\sigma(n)$
becomes trivial, and you can prune away large branches of non abundant
numbers (I'll let you work out the details).  Doing this efficiently makes a very good coding exercise.
The other opportunity for dramatic pruning comes from noting that if $n | m$ and
$m>n$ then $n/\sigma(n) > m/\sigma(m)$, and that if $(a,b)$ is an
amicable pair then $a/\sigma(a) + b/\sigma(b) = 1$ and that if $(a,b)$
is an amicable pair and $n | a$ then $\gcd(n,\sigma(n)) |b$.  So if we
let $m=\gcd(n,\sigma(n))$ then if $n/\sigma(n) + m/\sigma(m)<1$ no
number of the form $nk$ where $k$ is relatively prime to $n$ can be an
amicable number.  This seems like a lot of work, but abundant numbers
tend to have lots of small factors so most abundant numbers fail this test.  Heuristically it appears that the number of 'candidates' that remain is on the order of approximately $x/\log x$.
Using these techniques you should be able to compute
the pairs less than $10^{10}$ in under an hour on a fast machine (coding in C). 
A: Your list of the first 202 Amicable Number pairs... seems to have omitted the pair (356408, 399592).  If you check this pair, I think you'll find it belongs in the list of *203) pairs with the smallest number of the pair less than 66,000,000.
