In one of my courses of the summer, we came across the $3x+1$ problem. The function at hand is an arithmetic function of sorts; it is a function that both takes in positive integers and is recursive, and so calls itself (many) times before giving a final output. The function is:

(1)An example would be $f(11)$:

(2)So, 11 finally terminates at 1. The big (and unsolved) question is: does every positive integer terminate, i.e., does every input eventually reach 1?

It is easy to see that every $n = 2^a$ reaches 1 eventually, so in some sense, this is some measure of how close numbers are to powers of 2. As for determining whether the "$3x+1$ conjecture" is true, one method is to do some sort of strong induction argument: if $f(n) = f(k)$ for some $k < n$, then we say $f(n)$ terminates by our inductive hypothesis.

If we use this method, we don't need to look at any even numbers since $f(2m) = f(m)$. In addition, we don't need to examine numbers of the form $4m + 1$ since $3(4m + 1) + 1 = 12m + 4 = 4(3m + 1)$ and hence $f(4m + 1) = f(3m + 1)$ for $m > 1$. Thus, we've narrowed it down to numbers of the form $4x + 3$. However, this is still a rather harrowing task since there doesn't seem to be much of a pattern in the number of iterations (at least to me): $f(27)$ and $f(31)$ take over 90 iterations to dip below their starting points, but $f(35)$ takes less than 10.

Incidentally, there's a group in Portugal attempting to "prove" the $3x+1$ conjecture via computing "all" the possibilities. In September, they verified the conjecture for numbers through $19\cdot 2^{58} \approx 5.5 \cdot 10^{18}$. They also have done work on the similar $5x+1$ and $7x+1$ problems.