Is the Ratio of Associative Binary Operations to All Binary Operations on a Set of $n$ Elements Generally Small? I started thinking about the number of associative (binary) operations on a set with $n$ elements today.  Looking online I found this paper which indicates only $113$ of the possible $19,683$ operations on a three-element set satisfy association.  So, $50\%$ of binary operations on a two-element set satisfy association, and less than $0.6\%$ of all binary operations on a three element set satisfy association.  For an $n$-element set, where $n$ denotes a natural number, is the ratio $R_{n}=A_{n}/B_{n}$, of the number of associative binary operations $A_{n}$ to the number of all binary operations $B_{n}$, in general small?  I don't mean the following questions as equivalent, but since they seem more concretely answerable in principle, does 
$$
\lim_{n \to +\infty} \frac{A_{n}}{B_{n}} = 0 ?
$$     
Also, is $F : n \to R_{n}$ a monotonically decreasing function?
Some Background of the Question: In his Linear Algebra Problem Book 1995 on p. 6 Paul Halmos writes "The commonly accepted attitudes toward the commutative law and the associative law are different.  Many real life operations fail to commute; the mathematical community has learned to live with that fact and even to enjoy it.  Violations of the associative law, on the other hand, are usually considered by specialists only."  For all I know, Halmos might only have written that to motivate the study of Linear Algebra and doesn't quite literally mean what he appears to say.  But, if he means what he appears to say, and if $F : n \to R_{n}$ is monotonically decreasing, or $R_{n}$ is generally small, I think there's something amiss with what Halmos says, since non-associative operations seem so common that one may as well enjoy them.
 A: The number of associative binary operations on an $n$-element set is given at http://oeis.org/A023814
Perhaps you could follow the links at that page and report back to us. 
A: After getting a PhD at Har-cago, Allie got a job setting up the daily Binary Operation Table on the set $\{1,2,\dots, n\}$.
Every day Allie started by choosing $1 \circ 1$ at random, with all choices equally likely.  Whatever $1 \circ 1$ turned out to be, say $x$, Allie then chose at random the value of $x \circ 1$ among the legal choices. (Of course, if by luck it turned out that $x=1$, then $x \circ 1$ was already determined.)
It would be pointless to try to describe the rest of Allie's procedure, as it tended to change from day to day.  But the first two steps were always as described.
After the first two steps, Allie always checked whether or not the operation was (so far) associative. If the first choice was $1\circ 1=1$ (probability: $1/n$) we have automatic associativity, that is, associativity with probability $1$.
Otherwise (probability: $1-1/n$), if $1\circ 1=x \ne 1$, the operation is associative precisely if $x \circ 1=1\circ x$.  For any $x \ne 1$, this probability is $1/n$.  So the probability that $(1\circ 1)\circ 1=1\circ(1\circ 1)$ is
$$\left(\frac{1}{n}\right)(1) +\left(1-\frac{1}{n}\right)\left(\frac{1}{n}\right)$$
This simplifies to $(2n-1)/n^2$.
But this probability is $\ge A_n/B_n$, where $A_n$ and $B_n$ are defined as in the post.
Thus 
$$\frac{A_n}{B_n} \le \frac{2n-1}{n^2}$$
In particular, $A_n/B_n$ approaches $0$ as $n$ gets large.
The inequality we have obtained is tight at $n=1$, but absurdly weaker than the truth for large $n$. One could produce much improved estimates, and undoubtedly there is even a modest literature on the subject.
