We use the Erdős-Rado selection argument. Put
$$
m=R^{(2)}(k-1)+1.
$$
Starting with a sufficiently large reservoir, choose vertices $v_1,\ldots,v_m$ in order. After choosing $v_j$, successively halve the remaining reservoir for each $i<j$ so that the colour of
$$
\{v_i,v_j,x\}
$$
is constant as $x$ ranges over the final reservoir. The total number of halvings is at most $\binom m2$, so $2^{\binom m2}+m$ 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 $v_1,\ldots,v_{m-1}$, 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. Hence
$$
R^{(3)}(k)\le2^{\binom{R^{(2)}(k-1)+1}{2}}+R^{(2)}(k-1)+1.
$$
The <binomial upper bound for a Ramsey number> gives $R^{(2)}(k-1)<2^{2k}$, so the right-hand side is at most
$$
\boxed{2^{2^{Ck}}}
$$
for an absolute $C>0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 132 / 2 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-132/2/b/solution