User blog comment:ArtismScrub/how fast is this?/@comment-30754445-20171227173219/@comment-30754445-20171228062737

Yes, the new version seems to grow - indeed - as fω(n).

But my point is that you could have reached exactly the same growth rate without using the factorials at all, and it would have "cost" you only one more number inside the array.

In fact, you could have done it with exactly the same kind of rule that you used. Just replace every instance of "n![a,b]" with [n,a,b]:

k![n,a,b] = k![n,[n,b-1,c],c-1]

See? It's exactly the same kind of recursion. In both versions you are basically manipulating triplets of numbers (x,y,z) with a function F that follows the rule:

F(x,y,z) = F(x, F(x,y-1,z), z-1)

And it is this structure that gives your function most of its power.

The factorial doesn't add much. It just means that you get to fω(z+2) with x![y,z] instead of fω (z+1) with n![x,y,z].

BTW both versions can be extended to fωω(n)-level equally well.