User blog comment:Syst3ms/Breadth-Recursive Array Notation (BRAN)/@comment-25601061-20180424215049/@comment-5529393-20180425221959

Unfortunately, no.

A_0(n,n,...,n) with n entries is approximately f_ω (n).

A_1(n,n,...,n) takes A_0(n,n,...,n) (which is about f_ω (n)) and plugs that number into the number of entries of A_0(n,n,...,n), so we get about f_ω (f_ω(n)).

Similarly, A_2(n,n,...,n) takes A_1(n,n,...,n), which is about f_ω(f_ω(n)), and plugs it into the number of entries of A_2(n,n,...,n), so we get f_ω(f_ω(f_ω(f_ω(n)))).

More generally, if A_m(n,n,...,n) is about g(n), then A_{m+1}(n,n,...,n) will apply A_m twice, so it will be about g(g(n)). So, written as an iteration of f_ω's, the number of f_ω's will double each time you increase m. So A_m(n,n,...,n) is about (f_ω)^(2^m) (n), and z(n) = A_n(n,n,...,n) will be about (f_ω)^(2^n)(n), or about f_{ω+1}(2^n). So you've only gone one step up the fast-growing hierarchy.