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Given a positive integer $a$, the Ramsey number $R(a)$ is the least $n$ such that whenever the edges of the complete graph $K_n$ are colored using only two colors, we necessarily have a copy of $K_a$ with all its edges of the same color.

For example, $R(3)= 6$, which is usually stated by saying that in a party of 6 people, necessarily there are 3 that know each other, or 3 that don't know one another; but there is a party of 5 people without this property. This is probably known to everybody.

Slightly less known is the fact that any such coloring of $K_6$ in fact contains 2 monochromatic triangles.

The Ramsey multiplicity $m(a)$ (there does not seem to be a standard notation) is the largest number $m$ of monochromatic copies of $K_a$ that we can guarantee in any 2-coloring of $K_{R(a)}$. For example, $m(3)=2$, and Piwakowski and Radziszowski showed around 1999 that $m(4)=9$.

I have a couple of questions (please forgive me if they are trivial, I'm just beginning to form my intuitions in this field) :

  1. Is it known that $m(n)$ is monotonically increasing?

  2. Do we know anything about the rate of growth of the function $m(n)$?

I suspect that the answer to both questions is yes and that reasonable bounds for $m(n)$ are known, but haven't been able to locate any references. The best I know is that $$ m(n)\le \frac{\binom{r(n)}n}{2^{\binom n2-1}}, $$ (proved by Burr and Rosta in 1980), which is probably too high, and a recent result of Conlon suggests that $$ m(n)\ge C\frac{\binom{r(n)}n}{2^{n(3n-1)/2}} $$ for some appropriate $C$. I say "suggests" because Conlon's results carry some additional implicit constants that I haven't checked can be absorbed this way. (Please let me know if I am completely off the mark here.) [Edit: Unfortunately, Conlon's bounds (in his paper "On the Ramsey Multiplicity of Complete Graphs") do not apply here. No lower bound beyond $m(n)\ge1$ seems known.]

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Do you know $m(\omega)$, in the infinitary Ramsey case? –  Joel David Hamkins May 26 '10 at 17:06
    
Of course, in the case of $m(\omega)$, you would presumably add the condition that the monochromatic $K_{\omega}$ are maximal? –  Simon Thomas May 26 '10 at 17:22
    
@Joel: Isn't it clear that $m(\omega)=\omega$? For example we can partition the vertex set of $K_{\omega}$ into $\omega$ sets of size $\omega$. –  Tony Huynh May 26 '10 at 17:23
    
Even worse, $m(\omega) = 1$, since you can't do better than this if all the edges are given the same color. Pity about that! –  Simon Thomas May 26 '10 at 17:26
    
@Simon: If we insist that the monochromatic $K_{\omega}$ is maximal, isn't $m(\omega)=1$? I can colour all the edges red. –  Tony Huynh May 26 '10 at 17:26

2 Answers 2

up vote 9 down vote accepted

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.)

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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 $n \in \mathbb{N}$?

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Hi Simon; it is not obvious (to me). I've been trying to produce an argument to post here. Unfortunately, I am no longer sure about the tentative lower bound I mentioned in the question. –  Andres Caicedo May 31 '10 at 1:37
    
This makes your questions even more interesting! –  Simon Thomas May 31 '10 at 11:16

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