User blog comment:Triakula/Uncomputable functions defined as a sequence of computable functions/@comment-35470197-20200131231857/@comment-30279966-20200201055052

No, it doesn't. The growth rate of h(n) highly depends on the choice of the enumeration. However, I think that when the index i grows, we'll eventually find faster and faster-growing computable functions f_i(n). At least, I don't see how it is possible to order all computable functions so that growth rates of the functions are in decreasing or eventually non-increasing order. If it would turn to be possible, then h(n) is indeed computable.