User blog comment:AHAR/Clockwork Notation/@comment-1605058-20141109141031

This is actually quite far from what you claim - your first mistake is that actually $$^{^{^{^3...}}3}3=3\uparrow\uparrow\uparrow\text{tritri}=3\uparrow\uparrow\uparrow\uparrow 3$$. To be precise, we have $$^mn=n\uparrow\uparrow m,_mn=n\uparrow\uparrow\uparrow m,n_m=n\uparrow\uparrow\uparrow\uparrow m$$.