User blog comment:Simplicityaboveall/Insanely Fast-Growing Functions/@comment-1605058-20170915175750/@comment-5529393-20170915183157

Yeah, $$h(n,n)$$ is on the order of $$f_{\omega 2}(n)$$. Since $$H(n) < h^n(n,n)$$, it is on the order of $$f_{\omega 2 + 1}(n)$$.