User blog comment:Boboris02/Compearing BHAN to FGH/@comment-25601061-20161114222244/@comment-30118230-20161115163444

\(a\backslash^1_b = a(a(a(...(a)a)...)a)a)a\) with b+1 a's from the center out.

\(a\backslash^b_c =((...((a\backslash^b_{c-1})\backslash^b_{c-1})....)\backslash^b_{c-1}\)

\(a\backslash^b_1 = a\backslash^{b-1}_{a\backslash^{b-1}}\)