User blog comment:Primussupremus/Extending my notation past the 2nd level and into the 3rd./@comment-28606698-20170420181215/@comment-28606698-20170421104753

Look at your notation, up to 3-entry it works as Bowers's linear array since we use rather powerful rule 4 for defining of hyperoperations,

a,b,c={a,b,c}

but let compare cases for 4-entry array for your notation and Bowers's array: for Bowers's case all four entries are working, in your case only last entry is working, previous three entries are lazy, they just occupy space,

that is why

a,b,c#b≈{a,b,1,2}≈f_{w+1}(a)

a,b,c#(b)≈{a,b,2,2}≈f_{w+2}(a)

a,b,c#[b]≈{a,b,3,2}≈f_{w+3}(a)

Thus Bowers's array allows to reach w^2 using only four entries and full strenght of Bewers's array is w^w.