Proof of 4. It is fun.

Let S(n) be required.\

1. The weight start with 2^0 or 2^1 have the same S(n).

2. The first weight should be placed on left - first weight rule.

3. S(n) = 1.

4. n+1. Consider weight 1 (not 2^n that make thing compicated, and could scare people like me). It can be insert at any time and either side except the first weight rule. So, have 2n+1. The rests have S(n). Note that even weight 1 is added at the first time on left, it does not break the first weight rule of the rest n.

So, we have S(n+1) = (2n+1)S(n). Done.

Without knowing the result, I do not think I could solve it.