Conclusion: Plus a freaking constant!

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

Really, what does any of this have to do with powers of 2?

So I've been thinking about what I've doing as being working on the 2^a3^b problem, but in fact powers of 2 don't seem to enter much into it. I think I should be able to take this stuff (once I've properly proved the regularities in g_r(k), I mostly know how but I haven't done all of it - and actually I still haven't checked that they're good enough) and prove something of the form "For all numbers n not of the following forms, [list of forms], f(n)≥[something bigger than 3log_3(n)]". Depends how good the numbers are, though.

-Harry