Ramsey multiplicity

I emailed David Conlon about this question. He agreed to let me share his answer. In short, the problem very much seems to be open (I've added the relevant tag). As Thomas mentions, the upper bound I cite is straightforward. And nothing better is known!

If one looks for papers on Ramsey multiplicity, a few come up, but they deal with a different concept, that I explain below. The quotes are from Conlon's emails.

Unfortunately, the concept you're talking about is also known as the Ramsey multiplicity! There are very few references as far as I know. The only one I can think of offhand is the Piwakowski and Radziszowski paper which you quoted. Perhaps there's something in the references to that paper, but I doubt it somehow.

Indeed, in the papers I have seen (included P-R, where $m(4)=9$ is proved), there are no arguments about $m(n)$ for general $n$ (or even $n=5$).

The function you're interested in is rather amorphous, I'm afraid. My result will imply that if $n \ge 4^t$ you must have at least $n^t/2^{3t^2/2}$ copies of $K_t$ or thereabouts. But when your number is below $4^t$ it implies nothing.

In general, because we don't understand the Ramsey function, I find it hard to imagine how we might be able to say anything at all about $m(n)$. Unless there's an elementary argument which gives something. It reminds me of estimating the difference between successive Ramsey numbers like $r(n,n)$ and $r(n,n+1)$, where, though the difference is almost certainly exponential, the largest difference that can be guaranteed is tiny (I think linear or quadratic even, though I can't remember exactly).

Here, $r(m,n)$ are the usual Ramsey numbers (what I called $R(n)$ in the question, is $r(n,n)$ in this notation). In general, $r(m,n)$ is the smallest $k$ such that any coloring of the edges of $K_k$ with blue and red either contains a blue copy of $K_m$ or a red copy of $K_n$.

The only thing that appears clear to me is an upper bound following from the probabilistic method, namely $\displaystyle \frac{\binom{r(n)}{n}}{2^{\binom n2}}$. It's not even obvious how one would approach showing that the multiplicity is at least 2!

Finally, as to the question of how to call this concept:

I'd suggest that this be called the critical multiplicity or something like that, just to distinguish it from the usual multiplicity function.

The usual Ramsey multiplicity is defined as follows. It is significantly better understood than $m(n)$.

Let $k_t(n)$ be the minimum number of monochromatic copies of $K_t$ within a two-coloring of the edges of $K_n$, and let $$ c_t(n)= \frac{k_t(n)}{\binom nt} $$ be the minimum proportion of monochromatic copies of $K_t$ in such a two-coloring.

It is known that the numbers $c_t(n)$ increase with $n$. The Ramsey multiplicity of $t$ (or of $K_t$) is $\displaystyle c_t\lim_{n\to\infty} c_t(n)$.

(Relevant references can be found in Conlon's paper mentioned in the question.)

This is not an answer ... this is an even more trivial question. Why is it obvious with this definition of $m(n)$ that there doesn't exist a constant $k$ such that $m(n) \leq k$ for all $n \in \mathbb{N}$?