The number of non-isomorphic equivalence relations on a set of $n$ elements is the partition function
$$p(n) =\frac{1}{\pi\sqrt{2}} \sum\_{k=1}^{\infty} \sum\_{h=1}^{k} \delta\_{\gcd(h,k),1} \text{exp}\left(\pi i \sum\_{j=1}^{k-1} \frac{j}{k}\left(\frac{hj}{k} - \left\lfloor \frac{hj}{k} \right\rfloor - \frac{1}{2}\right) - \frac{2\pi i h n}{k} \right) \sqrt{k} \frac{d}{dn}\left[ \frac{\sinh\left(\frac{\pi}{k} \sqrt{\frac{2}{3}(n - \frac{1}{24})}\right)}{\sqrt{n - \frac{1}{24}}} \right]$$
The Hardy-Ramanujan asymptotic formula states that
$$p(n) \sim \frac{1}{4\sqrt{3}n}e^{\pi \sqrt{2n/3}}$$


By [this answer](https://math.stackexchange.com/questions/356995/counting-non-isomorphic-relations/357004#357004) (I would appreciate any reference to an actual derivation of this formula) the number of non-isomorphic relations on a set of $n$ elements is


$$a(n) = \sum\_{1s\_{1} + 2s\_{2} + \cdot\cdot\cdot =n} \left(2^{\sum\_{i,j \geq 1} \gcd(i,j)s\_{i}s\_{j}} \bigg/ \prod\_{k=1} k^{s\_{k}}s\_{k}!\right)$$


I have no idea about the asymptotics of $a(n)$, but if you know of a reference that would be amazing. My question is whether anyone has researched, or if you have any idea about, whether or not
$$\frac{p(n)}{a(n)} \sim 0$$
I conjecture that it is asymptotic to zero, but I have no idea how to prove it.



Asymptotics of the Number of Non-Isomorphic Equivalence Relations and the Number of Non-Isomorphic Relations
You don’t need either of the two fancy formulas. Since every equivalence relation is the kernel of a function from the $n$-element set into itself, their number is at most $n^n$ (and taking them up to isomorphism can only make it smaller). On the other hand, there are $2^{n^2}$ binary relations in total, and each isomorphism class has at most $n!$ elements, hence there are at least $2^{n^2}/n!$ nonisomorphic relations. Thus,
$$\frac{p(n)}{a(n)}\le\frac{n^nn!}{2^{n^2}}=2^{O(n\log n)-n^2}\to0.$$


