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Induction with Pascal's identity gives the binomial upper bound for a Ramsey number $R(s,t)\le \left(\genfrac{}{}{0ex}{}{s+t-2}{s-1}\right)$. Thus $R(s)\le \left(\genfrac{}{}{0ex}{}{2s-2}{s-1}\right)\le 4^{s-1}$, proving $R(s)=O(4^s)$.

For the lower bound, take a binomial random graph $G(n,1/2)$. The expected total number of $s$-cliques and independent $s$-sets isFor $n=\lfloor 2^{s/2}/2\rfloor$, this is at most $2^{1-s/2}/s!<1$ for all sufficiently large $s$. Some graph therefore has neither forbidden set, so $R(s)>n$. This proves $R(s)=\Omega (2^{s/2})$ by the probabilistic method.

The binomial upper bound for a Ramsey number gives $R(3,t)\le \left(\genfrac{}{}{0ex}{}{t+1}{2}\right)=O(t^2)$. For the triangle-free Ramsey lower bound by alteration, a binomial random graph $G(n,p)$ has expected triangle count $\left(\genfrac{}{}{0ex}{}{n}{3}\right)p^3$ and expected independent $t$-set count $\left(\genfrac{}{}{0ex}{}{n}{t}\right)(1-p{)}^{\left(\genfrac{}{}{0ex}{}{t}{2}\right)}$. Choose a realization whose sum of these counts is at most its expectation. Delete one vertex from each surviving forbidden set, iterating through the originally listed sets. The remaining graph has neither a triangle nor an independent $t$-set. This random alteration method provesFor large $t$, set $n=\lfloor (t/(4\log t){)}^{3/2}\rfloor$ and $p=n^{-2/3}$. The triangle term is at most $n/6$. Using $\left(\genfrac{}{}{0ex}{}{n}{t}\right)\le (en/t{)}^t$ and $1-p\le e^{-p}$, the logarithm of the independent-set term divided by $t$ is at mostwhich tends to $-\infty$. That term is eventually below $1$. Thus $R(3,t)>5n/6-1$ and

A graph Ramsey number $R(s,t)$ is the least $N$ such that every red-blue edge colouring of the complete graph $K_N$ has a red $K_s$ or a blue $K_t$. At a vertex in $K_{R(s-1,t)+R(s,t-1)}$, either at least $R(s-1,t)$ neighbours have red incident edges or at least $R(s,t-1)$ have blue incident edges. In the first case find a red $K_{s-1}$ and adjoin the vertex, or a blue $K_t$; the second case is symmetric. ThusStarting with $R(2,t)=t$ and $R(s,2)=s$, induction gives the binomial upper bound for a Ramsey numberFor the lower construction, partition $2t-2$ vertices into two classes of size $t-1$, colour inside-class edges red and between-class edges blue. Red cliques have at most $t-1$ vertices, and the blue graph is bipartite, so has no odd cycle.

Conversely, a graph without an odd cycle is bipartite: in each connected component colour vertices by the parity of the length of a path from a root; differing path parities would give an odd closed walk, hence an odd cycle. On $2t-1$ vertices one of the two blue bipartition classes has at least $t$ vertices. Every edge within it is red, giving a red $K_t$. Thus $2t-1$ is the exact threshold for this alternative.

The diagonal Ramsey number $R(s)$ is the least positive integer $N$ such that every red-blue edge colouring of the complete graph $K_N$ contains a monochromatic copy of $K_s$.

For the existence and bound, write $R(r,s)$ for the least $N$ forcing either a red $K_r$ or a blue $K_s$. Given a vertex $v$ of $K_{R(r-1,s)+R(r,s-1)}$, the pigeonhole principle says that at least $R(r-1,s)$ incident edges are red or at least $R(r,s-1)$ are blue. In the first case those red neighbours contain a red $K_{r-1}$, which joins to $v$ to make a red $K_r$, or a blue $K_s$; the second case is symmetric. HenceStarting from $R(1,s)=R(r,1)=1$ and using Pascal's identity gives the binomial upper bound for a Ramsey numberThereforewhich proves both existence and the required estimate.

To determine $R(3)$, color a cycle graph $C_5$ red and its complement graph, also a $C_5$, yellow. This colors $K_5$ without a monochromatic triangle, so $R(3)>5$. At any vertex of a red-yellow $K_6$, at least three of the five incident edges have one color. If an edge among their three other endpoints has that color, it completes a monochromatic triangle with the chosen vertex; if none does, those three endpoints themselves form a triangle of the other color. Thus $R(3)\le 6$, and

The same local argument proves uniqueness on $K_5$. No vertex can have three incident edges of one color: their other endpoints would either contain an edge of that color or form a triangle of the other color. Every vertex therefore has vertex degree at most two in each color, and because the two degrees sum to four, each is exactly two. Each color class is consequently a regular graph of degree two on five vertices, hence a single $C_5$. Up to relabelling, the coloring is exactly the cycle-and-complement coloring above.

Finally, the monochromatic-triangle counting formula gives, for a coloring of $K_6$ with red degrees $d_v$,The bound is attained by coloring all edges of $K_{3,3}$ red and the two complementary copies of $K_3$ yellow: there are no red triangles and exactly two yellow triangles. The least requested number is therefore

We use the Erdős-Rado selection argument. PutStarting with a sufficiently large reservoir, choose vertices $v_1,\dots ,v_m$ in order. After choosing $v_j$, successively halve the remaining reservoir for each $i<j$ so that the colour ofis constant as $x$ ranges over the final reservoir. The total number of halvings is at most $\left(\genfrac{}{}{0ex}{}{m}{2}\right)$, so $2^{\left(\genfrac{}{}{0ex}{}{m}{2}\right)}+m$ initial vertices suffice.

Colour the pair $ij$, for $i<j<m$, by the stabilized colour of $\{v_i,v_j,v_{\ell }\}$ for $\ell >j$. Among $v_1,\dots ,v_{m-1}$, the definition of $R^{(2)}(k-1)$ gives a monochromatic set of $k-1$ vertices. Adjoining $v_m$ gives a monochromatic $k$-set for the original triple colouring. HenceThe binomial upper bound for a Ramsey number gives $R^{(2)}(k-1)<2^{2k}$, so the right-hand side is at mostfor an absolute $C>0$.

The diagonal Ramsey number $R(t)$ is the least positive integer $n$ such that every red-blue colouring of the edges of $K_n$ contains a monochromatic $K_t$.

More generally, let $R(s,t)$ be the least $n$ forcing either a red $K_s$ or a blue $K_t$. We haveAssuming the two smaller numbers exist, colourand choose a vertex $v$. At least $R(s-1,t)$ of its incident edges are red, or at least $R(s,t-1)$ are blue. In the first case, the corresponding neighbourhood contains a red $K_{s-1}$, which extends with $v$ to a red $K_s$, or a blue $K_t$. The second case is symmetric. Thuswhich proves existence by induction.

Pascal's identity then gives the binomial upper bound for a Ramsey numberConsequently, for $t\ge 2$,