User blog comment:Simplicityaboveall/Insanely Fast-Growing Functions/@comment-28606698-20171025182658/@comment-30754445-20171103153821

Denis, note that Joe did diagonalize once.

The transition from h to H is exactly the kind of process you've described.

That's why H(11) is, indeed, comparable to fω+1(10) rather than f11(10).

From here on, however, H continues by simple recursion. There's no second diagonalization, so H(21) is comparable only to fω+11(10) rather than fω×2+1(10).