User blog comment:Boboris02/Boris´s sezration notation/@comment-29014792-20161005162834/@comment-30118230-20161006204437

Yes,but it's not the kind of nest you may be used to.

Remember:[n(@+@)n] expands into [n(@+[n(@+[...n(@+[n(@+n)n])n])n])n]...)n]

instead of [n([n([n(...[n(@)n])n])n]...)n] and even if you do all of that,there's gonna be @+(your result) in each step/floor.Also @+n works by apllying each previous step (@+(n-1)) and repeating it insanly many times,each one of which also breaks down into previous steps and so on...with @+@+@ each step gives you (growth rate)*w,so repeating the steps w times gives you more than w^(previous one).n*@ > w^(growth of (n-1)*@).This apllies for more powerfull notations of @ as well! And also you repeat the process way more than n(w) times.

That is another reason why functions using @ significantly overgrow similar expressions from the fast-growing hierarchy!