User blog comment:Lord Aspect/SATCLN IV/@comment-25601061-20180711040007/@comment-31966679-20180713030729

AAA[n] = AA[AA[n] - 1](n ^s here)AA[AA[n] - 1]

UU3[n] = A...(U3[n] A's in total)...A[n]

I may be wrong, because I was following whatever would work with the rules (because i wasn't the creator of this notation)