User blog comment:12AbBa/A Function Part II/@comment-31040770-20190913164047/@comment-39541634-20190915104053

Actually he did define nA(1,0) with the general rule, a bit later:

nA(@,x+1,0,O)=nA(@,x,nA(@,x,...nA(@,x,1,O)...,O),O)  with n A's.

Set O to empty, @ to empty and x=0 and you get:

nA(1,0) =nA(0,nA(0,nA(0,...nA(0,1)...)))) with n A's = nA(nA(nA(...nA(1)...)))

Which, curiously, means that nA(1,0)=nA(2). So that's a problem if you want A(1,0) to be an ω-level function, but it is most certainly well-defined.