Conclusion: Plus a freaking constant!

[sniffnoy]

So, does it generalize like I said? Oh, it generalizes alright. What this method ultimately tells me is that, pretty much, for all n except of certain exceptional forms, f(n)>3log₃(3/2n). In other words, I took the standard lower bound, and added a goddamned constant. Not even a large constant - 3log₃(3/2) is about 1.107. If you want to be more discriminating, you're occasionally instead adding 2+3log₃(3/4), or 4+3log₃(3/8), which are about 1.214 and 1.322, respectively. So yeah, this is just enough of an improvement on the standard lower bound to prove f(2^a3^b)=2a+3b for a≤19. But man, this is way suckier than I earlier realized. For a few days I thought I might have gotten a handle on the general case of 2^n, but that's basically the opposite of what is true.

-Harry