User blog comment:ReactorCoreZero/another ordinal question./@comment-37246647-20191128031538

I'll give you three answers, since ordinal hyperoperators are weird.

The first answer is that it's equal to epsilon-naught. This arises because by definition w^^(w+1)=w^e_0=e_0. The other two answers rely on us not following that logic.

The second answer, and the most common one, is that it's equal to phi(w,0). This is obtained through the standard method of ordinal hyperoperation, that is, to just use the same rules as CNF but extending them to the rest of the hyperoperators.

The third answer is that it's equal to phi(w,0,0). This is obtained through the "climbing" method of ordinal hyperoperation. If you look at w^(e_0+1), you might say that it's actually equal to w^(w^(w^w^w...)+1). If that +1 keeps on climbing up the tower, you'll find that w^w^w^...w w's...w^(w+1)=e_1, and that e_n=w^w^...w w's...w^(w+n). Following through with this logic, you'll find that w^^(w+1)=e_w^w, w^^(w*2)=e_e_0, w^^(w^2)=z_0, and that w^^^w=G_0, or phi(1,0,0). It will then follow that for n larger than 3, phi(n,0,0)=w^^^...n+2 arrows...^^^w, thus meaning that w^^...w arrows...^^w=phi(w,0,0).