About amicable numbers My question might seem wrong but I will explain it now:
We will define $\phi(a)$ as the sum of divisors of $a$ except $a$.
A pair $(a,b)$ is an amicable pair $\iff ((\phi(a) = b)  \land  (\phi(b) = a)) \land (a \neq b)$
Obviously if $(a,b)$ is an amicable pair $(b,a)$ is also an amicable pair. So, we will say $a$ is an amicable number $\iff$ There exist an $b$, such that $(a,b)$ is an amicable pair and $b < a$. Also we will say $a$ is a perfect number $\iff$ $\phi(a) = a$.
So my question is if we pick an arbitrary number, is it more likely an amicable number or perfect number? This question might be wrong because they might be finite but my question is simple: "Are there more amicable numbers than perfect numbers?"
I check with computer and believed that it is more likely an amicable number.
Let's formalize this as follows:  Let $P(n)$ be the number of perfect numbers $\leq n$, and let $A(n)$ be the number of amicable numbers less than $n$.  Does $\lim_{n\rightarrow\infty} \frac{P(n)}{A(n)}$ exist, and if so what is its value?
 A: I got the limits that I'm basing this on mainly from Wikipedia and WolframAlpha, so feel free to correct if something's wrong.
According to Wikipedia, $P(n)<c\sqrt n$, where c is a constant. I assume that this is a rough asymptotic limit, correct me if I'm wrong. It also says that $A(n)<ne^{-{\ln(n)}^{1/3}}$, so $\frac {P(n)}{A(n)}\sim\frac {c\sqrt n}{ne^{-{\ln(n)}^{1/3}}}$. The limit is easier to evaluate after taking the logarithm, so we have $\ln ({\frac {P(n)}{A(n)}})\sim\ln(n)^{1/3}-\frac{\ln(n)}2+\ln(c)$, so the limit is $-\infty$, since the $-\ln(n)\over 2$ dominates the function as n approaches $\infty$. To convert this back into its original form, we take $e^{-\infty}=0$. Therefore $\lim_{n\to \infty} \frac{P(n)}{A(n)}=0.$
A: EDIT: I misinterpreted the description of "amicable numbers" so my soft answer below is for the question of the relationship between numbers where $\phi(a) = \phi(b)$ to the perfect numbers.
This is a soft answer, but I'll express my thoughts here (slightly too long for a comment):
1) Every prime number (except 2) fits the above definition, since $\phi (p) = 1$ for $p$ = prime #. There are more numbers that fit this pattern ($25$, for example, $\phi(25) = 6 = \phi(6)$).
2) We know that the even perfect numbers have a 1-1 relationship with the Mersenne primes (see Wikipedia page for perfect numbers).
3) We don't know if there are any odd perfect numbers, but it does seem like if any do exist, there are many fewer of them than even perfect numbers.
4) So we may be able to limit the ratio of perfect numbers to numbers where $\phi(a)=\phi(b)$ by the ratio of Mersenne primes to primes overall. This ratio is another unknown, but empirical evidence suggests it's very small.
Again, a soft answer, but I enjoyed thinking about it.
