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I am not sure if "recurrence inequality" is the correct term or whether it is possible to actually find an answer to this problem but anyways.

Let $n$ be a fixed natural number. Let $R(x,y)$ be a symmetric function of two variables (the domain is the set of whole numbers), i.e. $R(x,y)=R(y,x)$ and further suppose that:

  • $R(x,y)\le R(x-1,y)+R(x,y-1)$

  • $R(0,i)=n$

Clearly $R(m,m)\le cn$ where $c$ is a natural number depending upon $m$. Is it possible to find the value of $c$ in terms of $m$?

I found that $R(1,1)\le 2n$ ; $R(2,2) \le 6n$ ; $R(3,3) \le 20n$ and $R(4,4) \le 70n$ by hand-computation. The pattern $2,6,20,70$ doesn't seem to be leading anywhere for me though. Does someone have an opinion concerning this?

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Is it really necessary to carry around the $n$? Why not divide everything by $n$ and pretend that $n=1$ all everywhere? – Srivatsan Dec 19 '11 at 12:33
$R(x,y)$ is the Ramsey number? – Paul Dec 19 '11 at 12:39
Yes indeed it is. – Shahab Dec 19 '11 at 12:49
up vote 2 down vote accepted

You can show easily by induction that $$R(x,y)\le \binom{x+y}y n.$$

If $x=0$ or $y=0$ this is clearly true.

Suppose this holds whenever $0\le x+y\le n$.

Let $x+y=n+1$ and $x,y>0$. Then you have $$R(x,y)\le R(x-1,y)+R(x,y-1) \le \binom{x+y-1}y n +\binom{x+y-1}{y-1} n=\binom{x+y}y n.$$

We have used Pascal's rule, see e.g. this question: Proving ${{n} \choose {r}}={{n-1} \choose {r-1}}+{{n-1} \choose r}$ when $1\leq r\leq n$

Note that I get $\binom21=2$, $\binom42=6$, $\binom63=20$, $\binom84=70$ in this way. I.e., I get 2,6,20,70 instead of 2,6,20,50. (I hope I did not make a mistake there.)

Now, when you mentioned in a comment that you are in fact interested in Ramsey numbers, this inequality $$R(r,s) \leq \binom{r+s-2}{r-1}$$ seems to be well-known, it is mention in wikipedia article on Ramsey numbers.

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Thanks. I was aware of this inequality but am not sure is relevant here. – Shahab Dec 19 '11 at 13:26

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