h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

If we are trying to proof 3n+1, we only need to show that each number eventually is mapped to a smaller number.

If n is even, then f(n) (f is 3n+1 if n is odd, n/2 if n is even) is less than n.
If n is 1 mod 4, let n=4m+1. Then f(n)=12m+4 and f(f(f(n)))=3m+1, which is less than n.
This leaves n=3 mod 4. Here, f(f(n)), where n=4m+3, is 6m+5, which is odd, so f(f(f(f(n)))) is 9m+8. Here, if m is a multiple of 4, giving n=3 mod 16, then 3n+1 holds.

This case-by-case reasoning can go on and on; I have done it to taking n=256m+k. However, this will never arrive at a proof because for n=(2 ^ h)*m + (2 ^ h) -1, you must make an assumption on the parity (even or odd) of m, which further bifurcates the number of cases. Also, it is thus impossible to make a claim that the iterations of n never exceed a fixed multiple of n, because 2^k-1 iterates to 3^k-1.

I hope this clears up some of the confusion, if you understand what I said. If you didn't, the problem is confusing, so don't be worried.

1+8/8+0+5*8

*blinks*

peace.

That's good reasoning, Researcher! I did the same tests myself for every number of form 1024k + n. So after a lot of hard work, I couldn't realy conclude anything.