[10intheCrunch]

OK I tried to do some of the first one, and I didn't get around to an actual proof yet, but as it's late and I'm really tired I'll try and post what I found out.

First thing I realized when iterating the function through itself was that f(3^m * n) = 2^mf(n) where m is any nonnegative interger. That kinda helps down the road.

I started out just by listing some function values, 6-->12. I called f(3)=x, so f(4)=x+1, f(5)=x+2. f(6) I called y, so f(7)=y+1, f(9)=y+2. f(9)=2x, f(10)=2x+1, f(11)=2x+2 and f(12)=2x+2. I noted that f(11) = f(12) = 2f(4), thought it might be general, so that f(3n-1) = f(3n).

About this time I wanted to link up x and y so I could compare them directly, and the magic of zero jumped out and smacked me in the head (couldn't believe I hadn't thought of it first). f(0) = z, f(1) = z+1 = x/2 (from f(3)), f(2) = y/2 (from f(6)). So, x/2 + 1 = y/2 => x + 2 = y. This checks out my f(3n-1_ = f(3n) as f(6) = y = x + 2 = f(5). However, it fails at f(17) = 2x + 6 != f(18) = 2y = 2x + 4.

So then I scrolled down as I type this and realized that n > 0 is a check on the system. As a Q to the writer, does this mean that 0 is not in the domain of f? "Nonnegative integers" would indicate that 0 is allowable.

I wrote out a little list of integers and the laws seem to break down at some point, or at least stop agreeing with each other.

I think I might have gotten in over my head here. I'll try again tomorrow...nice hard problem (at least for those of us who suck at proofs...).