User blog comment:Wythagoras/Upper bound on the Buchholz hydra/@comment-3427444-20140505134038/@comment-5150073-20140505134956

I guess that $$BH(n) < f_{\psi(\psi_1(\psi_2(\cdots(\psi_{n-3}(3))\cdots)))}(6)$$. So:

$$BH(3) < f_{\psi(3)}(6) = f_{\varepsilon_3}(6)$$

$$BH(4) < f_{\psi(\psi_1(3))}(6)$$

$$BH(5) < f_{\psi(\psi_1(\psi_2(3)))}(6)$$