I read that the interesting identity $R_3(k) > \frac{k}{3e} \sqrt {3}^k$ and $R_r(3) > 2^r$ holds (where R stands for the Ramsey numbers) but no proof was given. I'm not sure how to go about the proof. How is this done?
2026-03-29 08:57:55.1774774675
Show that $R_3(k) > \frac{k}{3e} \sqrt {3}^k$ and $R_r(3) > 2^r$ .
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