User blog comment:Alejandro Magno/My -illion is bigger!/@comment-4297047-20140923011640/@comment-25371339-20140923011932

Ok, here's a big one.

Exihyperillion = R(16)

where R(n) = H(H(H(...(H(H(H(n))))...))) with n levels

This is still valid, because R(n) can be represented with the H Function.