User blog comment:Simplicityaboveall/Extremely Large Numbers 3/@comment-24920136-20160803133009/@comment-5529393-20160803224914

Chronolegends is correct. Just look at the definitions; heshbar-n is the heshbar function iterated n times, just as f_{a+1}(n) is the f_a function iterated n times. So if the heshbar function is comparable to f_a, then heshbar-n will be comparable to f_{a+1}(n).