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 $b &lt; a$. 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?
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, $P(n)&lt;c\sqrt n$, where c is a constant. I assume that this is a rough asymptotic limit, correct me if I'm wrong. It also says that $A(n)&lt;ne^{-{\ln(n)}^{1/3}}$, 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 $\lim_{n\to \infty} \frac{P(n)}{A(n)}=0.$
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, $P(n)&lt;c\sqrt n$, where c is a constant. I assume that this is a rough asymptotic limit, correct me if I'm wrong. It also says that $A(n)&lt;ne^{-{\ln(n)}^{1/3}}$, 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 $\lim_{n\to \infty} \frac{P(n)}{A(n)}=0.$