User blog comment:TechKon/Some "function" i made/@comment-24923514-20181211044745

This kind of reminds me of the process for Graham's number (the popular version): g_(n+1) = 3^^...^^3 with g_n up arrows. Obviously your function has E*(n) for the two arguments where the g-terms have threes, but this distinction becomes diminishingly important compared to the number of up arrows. And of course you use hyperoperations instead of up arrows, so the kind of operations differ by 2 levels, but again this matters less and less when the level of hyperop becomes large. Since g_n is on par with f_(ω+1)(n), I'd think that E* is on roughly the same level. 👍