Cocycles and the Collatz problem?

Question

Let $T(n) = n+R(n)$, where $R(n) = -n/2 $ if $n\equiv 0 \mod 2$ else $R(n) = \frac{n+1}{2}$. $R(n)$ is the Cantor ordering of the integers:

https://oeis.org/A001057

In the Collatz problem, one is interested in the parity vector $x(n)$ of $n$:

$$x_k(n) : = T^k(n) \mod(2)$$

It is easy to see, that iterations of $R$ map each integer to either $0=R(0)$ or $1=R(1)$.

2-Cocycles:

Define

$$\epsilon_k(a,b):= T^k(a+b)-T^k(a)-T^k(b)\mod(2)$$

Then $\epsilon_k: \mathbb{Z}\times \mathbb{Z} \rightarrow \mathbb{F}_2$ is a 2-cocycle for each $k \in \mathbb{N}_0$:

This can be proven because the addition in $\mathbb{Z}$ is associative:

$$T^k(a+(b+c)) = T^k((a+b)+c)$$

1. Question:

Is it true that for $k \ge2$:

$$T^k(n) \equiv \sum_{i=0}^k \binom{k}{i} R^i(n) + \sum_{i=1}^{k-1} \sum_{j=0}^{k-i-1} \epsilon_i(R^j(n),R^{j+1}(n)) \binom{k-i-1}{j} \mod(2)$$?

Edit 07.07.2024: This can be proven by induction on $k$.

On the right hand side, we see only iterations of $R$, denoted as $R^i,R^j$. But then for large enough $k$ we have $R^l(n) = R^k(n) = 0 \text{ or } 1$ for $l \ge k$. Hence maybe this might give a hint as to why the parity vector of each natural number repeats with $0,1$ after some time (Collatz conjecture).

I am not saying that it proves the Collatz conjecture.

2. question: What other applications of cohomology could be applied in this situation?

Here is some Sagemath Code for experimentation:

def TT(m):
    if m%2==0:
        return m//2
    else:
        return (3*m+1)//2
    
def R(n):
    if n%2==0:
        return -n//2
    else:
        return (n+1)//2

def iterFunc(func,k,n):
    if k==0:
        return n
    else:
        return func(iterFunc(func,k-1,n))
    
    
def Tk(k,n):
    return iterFunc(TT,k,n)

def Rk(k,n):
    return iterFunc(R,k,n)

def eps(a,b,base=2,k=1):
    return (Tk(k,a+b)-(Tk(k,a)+Tk(k,b)))%base
    

def rhs(k,n):
    if k>=2:
        return (sum(binomial(k,i)*Rk(i,n) for i in range(k+1))+sum(sum(eps(Rk(j,n),Rk(j+1,n),base=2,k=i)*binomial(k-i-1,j) for j in range(k-i-1+1)) for i in range(1,k-1+1)))%2
    

for n in range(1,100):
    for k in range(2,15):
        print(n,k,Tk(k,n)%2==rhs(k,n))

Edit:

Let $I(n)$ denote the stopping time when the starting number $n$ iterates to either $0$ or $1$ under $R$:

$$I(n) := \min_{R^i(n) \in \{0,1\}} \{i\}$$

Then by empirical observation for $n=1,\cdots,10^5$ I think that:

$$T^{I(n)^3}(n) \in \{1,2\}$$