Ruby, 100 bytes, score = fω³-1(n) Basically borrowed from my Largest Number Printable, here's my program: ->n{k=0;f=->a,b=k,c=k{k.times{k*=c>0?f[a,b,c-1]:b>0?f[a,b-1]:a<0?k:f[a-1]}};k+=1 while f[k]<n;k} Try it online Basically computes the inverse of fω³(n) in the fast growing hierarchy. Th...