Puzzle time - binaries

[Klaus]

Consider a string of binary numbers s, such as s = 1011.

Now consider a machine that works on s. It performs the following three steps:

1. Append 1 to the right end of s.
2. Add the new s from step 1 to the old s with binary addition
3. Remove all trailing 0s.

The machine iterates these steps and stops only when s = 1.

For instance, for s = 1011 we have s = 10111 after step 1, s = 10111+1011 = 100010 after step 2, s = 10001 after step 3.

After the next iteration we get s = 01101, then 0101, then 1. The machine stops.

Can you find an s such that the machine doesn't stop? Or conversely, can you make an argument why the machine will always stop for any s?

[jon-nyc]

I think it will always Collatz to 1 in the end. But that's just a conjecture.

[jon-nyc]

Kind of a dick move, by the way

[Klaus]

Damn, my superficial masquerade of the problem failed.