Q:

Preserving historical information of the Collatz function?

In some sense this two equations are the same, namely $f_2$ preserves the historical information of $f_1^n$, where the exponent is function composition, but I am not sure how to show this rigorously. $f_1$ is well known,
$
f_1(s)=
\begin{cases}
 3s+1&\text{if}\, s \in \mathbb{N}_{odd}\\
 \frac{S}{2}&\text{if } s \in \mathbb{N}_{even} \\
\end{cases}
$ 
$f_2(s)=3s+2^{E(s)}$
where $E(s)=\max(\; b \in \mathbb{N} \cup \{0\}: \;\;2^b|s\;)$, and $$f_1^n(s)=1 \iff (f_2^m(s) \text{ is a power of two for some } m)$$
How do I show $f_1$ and $f_2$ are equivalent in this sense? Any ideas or references would be appreciated. 
Edit 1:
My argument to move from $f_1$ to $f_2$ is as follows:
Represent the input $s$  in binary form. And consider $f_1^{odd}(s)=3s+1$ and $f_1^{even}(s)=\frac{s}{2}$ for odd and even inputs, respectively.
Applying $f_1^{even}$ repeatedly on the input $s_{even}$ until we obtain an output that is $s_{odd}$ to then use $f_2^{odd}(s)$ amounts to two operations:

Finding the first/lowest bit set on $s_{even}$ and adding one bit at that position. This is equivalent to adding $2^{E(s_{even})}$ to $s_{even}$; this will make that bit position become 0.
Adding $(s_{even}<<1)$ to the above computation, i.e. shifting by one to left or multiplying by two. Leaving the position where the lowest bit was set on $s_{even}$ intact. 

Hence, after doing the two operations above we get, $f_2$. In particular, at all times 
$$ popcount(\; f_{odd}(s)  \;) = popcount(\;f_2(s) \;) $$
$$ popcount(\; f^n_{even}(s) : n = \text{ (a power until output is odd)} + 1  \;) = popcount(\;f_2(s) \;) $$
The bits that change after both operation are the same bits modulo the number of trailing zeroes; that is, in $f_2$ they are at an offset.
Hence, when the Collatz function has a one bit output, our function $f_2$ is power of two; as only powers of two have only one bit set in binary.
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Edit 2: By convention we use $\mathbb{N}$ to mean natural numbers greater than zero.
Consider the following functions:
\begin{align}
E(s) &= \max(\; r \in \mathbb{N} \cup \{0\} \; : \; \; \; 2^r\; | \;s) \\
C(s) &=
\begin{cases}
 3s+1&\text{if}\, s \in \mathbb{N}_{odd}\\
 \frac{s}{2}&\text{if } s \in \mathbb{N}_{even} \\
\end{cases}\\
G(s) &= C^{E(s)+1}(s) \\
F(s) &= 3s + 2^{E(s)} 
\end{align}
By convention we define $G^0(s)=s,\;F^0(s)=s$. We make the following claims:
$\begin{align}
 1.& \quad \forall s \in \mathbb{N},&  2^{E(s)}G(s) = F(s) \\
 2.& \quad \forall n,s \in \mathbb{N}, &\left( 2^{\sum_{k=1}^n E(G^{k-1}(s))} \right) G^n(s)=F^n(s)\\ 
\end{align}$
$
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$
Claim 1 Proof
Suppose $s=2^ks_0$ where $s_0$ is odd. We have $E(s)=k$. We start as follows,
\begin{align*}
G(s) &= C^{E(s)+1}(s)          \\
G(s) &= C^{k+1}(s)             \\
G(s) &= C^{k+1}(2^ks_0)   \\
\end{align*}
since $2^ks_0$ is even until it is divided by $2$ a total number of $k$ times, we have  
\begin{align}
G(s) &= C^{1}(s_0)   \\
G(s) &= 3s_0+1   
\end{align}
Hence, 
\begin{align}
2^{E(s)}G(s) &= 2^{E(s)}( 3s_0+1) \\
             &= 2^k(3s_0+1)      
\end{align}
Now we calculate $F(s)$,
\begin{align*}
F(s) &=  3s+2^{E(s)} \\
     &=  3s+2^{k} \\
     &=  3(2^ks_0)+2^k      \\
     &=  2^k(3s_0+1)      \\
\end{align*}
Therefore, we have $  2^{E(s)}G(s) = F(s) $. 
A proof for claim 2 would get the points.

A:

$\text{let } {x=E(s)} \text{ therefore } {\exists\space y\in\Bbb{N}\space}\text{|}{\space 2^xy=s}\text{ (This implies } y \text{ is odd).}$
$\text{Also let }{f_3(s)=3(\frac{s}{2^x})+1=(3y+1)}\text{.}$
${f_3\left(s\right)}\text{ is equivalent to using } {f_1}\text{ on } s \text{ until an odd step is applied.}$
$\text{So }$
$\exists\space p\space \text{|}\space f_1^p(s)=1\iff\exists\space q\space|\space f_3^q(s){=4\text{ [1]}}$ 
${f_3^q(s)=4 \iff f_3^{q-1}(s)=2^k} \text{ (for some k}{\in\Bbb{N})\space [2]}$
$\text{If we apply one step of } {f_2} \text{ on } {s} \text{ we have: }$
${f_2(s)=f_2(2^xy)}$ ${=}$ ${3\cdot 2^xy+2^x=2^x(3y+1)}$
${f_2(s)} \text{ is a power of two } {\iff (3y+1)} \text{ is a power of two.}$ 
$\text{Let} {f_4(s)=\frac{f_3(s)}{2^x}=3y+1}\text{.}$
$\text{So } {f_4(s)} \text{ is a power of two } {\iff f_2(s)} \text{ is a power of two.[3]}$
${f_4(s)=(3y+1)=f_3(s)} \text{ implies  that } {f^n_4(s)=f^n_3(s)}\text{.[4]}$
$\text{The logical connection between Statements [1]-[4] shows that: }$
${f_1^n(s)=1 \iff (f_2^m(s) }\text{ is a power of two for some m)}$
$\text{Recommended ordering [1][2][4][3] or [3][4][2][1]}$
Edit to prove claim 2:
$\text{we want to show by induction that:}$
$$\left( 2^{\sum_{k=1}^n E(G^{k-1}(s))} \right) G^n(s)=F^n(s)\Rightarrow \left( 2^{\sum_{k=1}^n E(G^{k}(s))} \right) G^{n+1}(s)=F^{n+1}(s)$$
$\text{(we already have the base case from claim 1)}$
$\text{I first want to prove an identity of the } F \text{ function. }(F(\frac{s}{2^c})=\frac{F(s)}{2^c})$
$\text{notice that } E(\frac{s}{2^c})=E(s)-c \text{.}$
$F(\frac{s}{2^c})=\frac{3s}{2^c}+2^{E(\frac{s}{2^c})}=\frac{3s}{2^c}+2^{E(s)-c}=\frac{3s+2^{E(s)}}{2^c}=\frac{F(s)}{2^c}$
$$\left( 2^{\sum_{k=1}^n E(G^{k-1}(s))} \right) G^n(s)=F^n(s)$$
$$ G^n(s)=\frac{F^n(s)}{\left( 2^{\sum_{k=1}^n E(G^{k-1}(s))} \right)}\text{ [0]}$$
$$ G^{n+1}(s)=G\left(\frac{F^n(s)}{\left( 2^{\sum_{k=1}^n E(G^{k-1}(s))} \right)}\right)\text{ [1]}$$
$\text{On the right side of [1] the input of G is divided by } 2^{E(G^n(s))}$
$\text{ then is multiplied by three then summed with one as shown in [2]}$
$$ G^{n+1}(s)=\frac{3F^n(s)}{\left( 2^{\sum_{k=1}^n E(G^{k}(s))} \right)}+1 \text{ [2]}$$
$$ 2^{E(G^n(s))}G^{n+1}(s)=\frac{3F^n(s)}{\left( 2^{\sum_{k=1}^n E(G^{k-1}(s))} \right)}+2^{E(G^n(s))}\text{ [3]}$$
$\text{the right side of [3] is equivalent to plugging in the right side of [0] in F}$
$\text{ as shown on the right side of [4]}$
$$ 2^{E(G^n(s))}G^{n+1}(s)=F\left(\frac{F^n(s)}{\left( 2^{\sum_{k=1}^n E(G^{k-1}(s))} \right)}\right)\text{ [4]}$$
$\text{using the F identity}$
$$ 2^{E(G^n(s))}G^{n+1}(s)=\frac{F^{n+1}(s)}{\left( 2^{\sum_{k=1}^n E(G^{k-1}(s))} \right)}\text{ [5]}$$
$\text{finally}$
$$\left( 2^{\sum_{k=1}^n E(G^{k}(s))} \right) G^{n+1}(s)=F^{n+1}(s)$$
Edit2 shortened proof of claim 2:
$\text{we want to show by induction that:}$
$$\left( 2^{\sum_{k=1}^n E(G^{k-1}(s))} \right) G^n(s)=F^n(s)\Rightarrow \left( 2^{\sum_{k=1}^n E(G^{k}(s))} \right) G^{n+1}(s)=F^{n+1}(s)$$
$\text{(we already have the base case from claim 1.) Claim 1 also shows that:}$
$$F(z)=G(z)2^{E(z)}\text{ (for any z }\in\Bbb{N}\text{) [1]}$$
$\text{I first want to prove an identity of the } F \text{ function. }(F(s2^c)=F(s)2^c)$
$\text{notice that } E(s\cdot 2^c)=E(s)+c \text{.}$
$F(s2^c)=3s2^c+2^{E(s2^c)}=3s2^c+2^{E(s)+c}=(3s+2^{E(s)})2^c=F(s)2^c$
$$G^n(s)\left( 2^{\sum_{k=1}^n E(G^{k-1}(s))} \right)=F^n(s)$$
$$F\left(G^n(s)\left( 2^{\sum_{k=1}^n E(G^{k-1}(s))} \right)\right)=F^{n+1}(s)$$
$\text{By the F identity}$
$$F\left(G^n(s)\right)\left( 2^{\sum_{k=1}^n E(G^{k-1}(s))} \right)=F^{n+1}(s)\text{ [2]}$$
$\text{By using [1] when } z=G^n(s),$
$\text{the } F(G^n(s))\text{ term in [2]  
can be substituted for } G^{n+1}(s)2^{E(G^n(s))}\text{.}$
$\text{Which gives the result:}$
$$G^{n+1}(s)\left( 2^{\sum_{k=1}^n E(G^{k}(s))} \right)=F^{n+1}(s)$$

