User blog comment:LittlePeng9/FORMAL comparison of Hydra battles and Hardy hierarchy/@comment-5529393-20141016225656

You can see the value of the Hydra function in terms of the modified Hardy Hierarchy directly:  Define H_a(n) to be the final index when the Hydra is killed, rather than the number of steps needed to kill it. Then it is clear than H_0(n) = n, H_{a+1}(n) = H_a(n+1), and H_a(n) = H_{a(n)}(n+1), so the Hydra hierarchy is exactly the modified Hardy hierarchy. Of course the final index is n more than the number of steps, so that gives is H_a(n) = L_a(n) + n without induction.

This modified Hardy Hierarchy is what gives us the bounds in Kirby-Paris Hydra article. Letting H' stand for the modified Hardy Hierarchy and H stand for the original, we have for example:

$$Hydra(4) = H'_{\omega^{\omega^\omega}}(2) = H'_{\omega^{\omega^2}}(3) = H'_{\omega^{\omega 3}}(4) = H'_{\omega^{\omega 2 + 4}}(5) > H_{\omega^{\omega 2 + 4}}(5) = f_{\omega 2 + 4}(5)$$