Trying This

As a problem solving exercise, I've started working on the problem in /TryThisProblem on Danny's problem-solving wiki site (https://wwwcgi.cdf.toronto.edu/~heap/cgi-bin/Solvent/wiki.pl?Problem_Solving_Home_Page/TryThisProblem).

I like this problem. As Danny says, good problems fight back. This problem, to me, seems to have a challenging depth to it. It doesn't immediately present me with a fairly clear path of attack. But, it did immediately get me thinking, and really grabbed hold of me.

My first step was to (obivously) read the problem. As I was doing this for the second time, I was noting the problem down on paper, more in my own words. For instance, I defined f(n) more symbolically than is done on the wiki page.

This was done to help me frame the problem, and get the problem set up to use a the "mathematical vocabulary" (if you will) that fits me best.

As I was looking at this problem, the way I had it laid out helped. Examining it, I started thinking a couple main ideas:

• Couldn't a closed form for F(m, n) be found?
  
• Would this hypothetical closed form even help with the main thrust of the problem?
• This "closed" form should have two cases, one for each parity of input n (i.e. one for an even n, one case for an odd n)
• It would be nice to be able to look at the results of each application of f(n) lying underneath F(m, n). The parity of these sub-results, in particular, seemed like a key to a closed form, or at least getting insight into the behaviour of F(m, n)

Trying to start on the road to this closed form, armed with my two cases of input n's, I looked at the results of a single application of f(n), given generalized input n's (one for each case, 2k for even, 2k-1 for odd, where k is in the natural numbers, an k > 0).

With even this starting examination, I felt a wave of insight wash over me. The generalized behaviour of compositions of f(n) started to become quite a bit more clear.

My initial results:

• In the case that n is even, f(n) is either even, or odd. In other words, f(n) produces a natural number greater than 0.
• In the case that n is odd, f(n) is always even. This result was intriguing to me.

So, if n is odd, f(n) then produces an even result. This result, if passed to f(n) again, is subject to the arithmetic of the even branch of f(n). This even branch then produces either an even or odd result.

Now it seems to me that this either even or odd result of f(n) from an even input n can be passed to f(n) again, over and over until f(n) produces an odd result. Passing this result to f(n) again sends it to the odd branch, which produces an even number. Evidently, another application of f(n) to this number starts the process described above again.

The way I think of it is that f(n) takes odd numbers and makes them even, which "bounces" them back to the even branch. This branch then reduces them 1 (the desired condition), or reduces them to some odd number greater than 1. The odd branch makes the number even then "bounces" it back to the even branch. So it appears that we have this interplay, a numerical game of catch between the two branches of the function under repeated the composition used in F(m, n).

So, a closed form isn't so simple, but poking around in the area of that problem at least gave me some somewhat more refined ideas about the main problem.

And that's as far as I've gotten. I know it's not much, but it feels like this exercise was already a valuable experience. And hopefully what I've written above doesn't come across as rambling overmuch. What I described was the thought process I went through.

If I can, I will make another post if I make progress on this problem.