How to prove this equation related to the Collatz conjecture (residue)

Question

Consider this formula:

$2^{E(N)}=N\cdot 3^{O(N)}\cdot RES(N)$.

This is a formula, it really works for the 3x+1 problem. I found it in E.R. site. E(N) and O(N) are the number of even steps and odd steps in the sequence.

N is the integer from where you start the sequence.

RES(N) is the so called residue of N and it is the product of $(1+\frac{1}{3\cdot S_i})$ taken over all the odd $S_i$ in the sequence.

So for example if you strat form N=5, O(N)=1 because there is only the term 5 which is odd in the sequence, whereas E(N)=4, because there are four even terms in the sequence starting form 5: 16, 8, 4, 2.

So we have:

$2^4=3\cdot 5\cdot (1+\frac {1}{15})$

From N=3 we have O(N)=2, E(N)=5 and so

$2^5=3\cdot 3^2\cdot (1+\frac {1}{15})\cdot (1+\frac {1}{9})$

My question is:

how to prove this formula more generally?

If there is something wrong, please tell me.

Answer 1

With $S_0=N$ $$\frac{3S_0+1}{2^{\nu_2(3S_0+1)}}=S_1$$ can be rewritten as $$(3+\frac{1}{S_0})=2^{\nu_2(3S_0+1)}\frac{S_1}{S_0}$$ Now you have

$(3+\frac{1}{S_0})=2^{\nu_2(3S_0+1)}\frac{S_1}{S_0}$

$(3+\frac{1}{S_1})=2^{\nu_2(3S_1+1)}\frac{S_2}{S_1}$

...

$(3+\frac{1}{S_{O(N)-1}})=2^{\nu_2(3S_{O(N)-1}+1)}\frac{S_{O(N)}}{S_{O(N)-1}}$

You multiply every LHS/RHS to get

$$3^{O(N)}(1+\frac{1}{3S_0})(1+\frac{1}{3S_1})...(1+\frac{1}{3S_{O(N)-1}})=\frac{S_{O(N)}}{S_0}\prod\limits_{k=0}^{O(N)-1}2^{\nu_2(3S_k+1)}$$ and with $S_{O(N)}=1$

$$N\cdot 3^{O(N)}\prod\limits_i(1+\frac{1}{3S_i})=2^{E(N)}$$