I was reading "On some sequences of integers" by Erdös and Turán, and was having some trouble following one of the arguments. You can find the paper here
Right after the proof of Theorem 2, they say:
By a similar but very much longer argument we find that $$r(18)=r(19)=r(20)=8.$$ On the other hand, $r(21)=9$, since $1,3,4,8,9,16,18,19,21$ is an $A$ sequence; further, $$r(22)=r(23)=9.$$ Hence, as previously, we find that, for sufficiently large $N>N(\epsilon)$, $$r(N)<\left(\frac38+\epsilon\right)N.$$
But I don't really understand why this implies the $\frac38$ bound. I'm assuming you want to say something about $r(2\cdot 21+1)$ or something, using the fact that $r(m+n)\le r(m)+r(n)$, but I just don't see it.
Any help is much appreciated, thanks!
My guess is that he does induction with 23 the same way he did with 17, getting:
$$ r(23) = 9 \Rightarrow r(2^4 + 2^3 - 1) = 9 \Rightarrow r(2^{k+4} + 2^{k+3} - 1) \leq 9 * 2^k $$
Then, we have:
$$ \frac{r(2^{k+4} + 2^{k+3} - 1)}{(2^{k+4} + 2^{k+3} - 1)} \leq \frac{9 * 2^k}{(2^{k+4} + 2^{k+3} - 1)} = \frac{3}{8 - 2^{-k}3^{-1}} \sim \frac{3}{8} + \epsilon $$