Is this function definition related to the Collatz conjecture valid?

Question

Is the following function definitions related to the Collatz Conjecture valid? Is there a simpler or more standard way to define $C_n(x)$?

Here's what I did:

Let:

• $C(x) = \dfrac{3x+1}{2^w}$ where $w$ is the highest power of $2$ that divides $3x+1$

Goal:

• Define $C_n(x)$ where:

$$C_n(x) = C_1(C_2(C_3(\dots C_n(x)\dots)))$$

• Define $w_i$ as the highest power of $2$ that divides $C_i(x)$:

Claim:

$$C_n(x) = \frac{3^n x + 3^{n-1} + \sum\limits_{i=1}^{n-1}3^{n-1-i}2^{\left(\sum\limits_{k=1}^{i}w_k\right)}}{2^{\left(\sum\limits_{j=1}^{n} w_i\right)}}$$

Argument:

(1) Base Case: $n=2$: $C_2(x) = C(C(x)) = C\left(\dfrac{3x+1}{2^{w_1}}\right) = \dfrac{3\left(\frac{3x+1}{2^{w_1}}\right)+1}{2^{w_2}} = \frac{3^2x + 3 + 2^{w_1}}{2^{w_1 + w_2}}$

(2) Assume that it is true up to $n$ so that:

$$C_n(x) = \frac{3^n x + 3^{n-1} + \sum\limits_{i=1}^{n-1}3^{n-1-i}2^{\left(\sum\limits_{k=1}^{i}w_k\right)}}{2^{\left(\sum\limits_{j=1}^{n} w_i\right)}}$$

(3) Then:

$$C_{n+1}(x) = C(C_n(x)) = \frac{3\left(\frac{3^n x + 3^{n-1} + \sum\limits_{i=1}^{n-1}3^{n-1-i}2^{\left(\sum\limits_{k=1}^{i}w_k\right)}}{2^{\left(\sum\limits_{j=1}^{n} w_i\right)}}\right)+1}{2^{w_{n+1}}} = \frac{3^{n+1} x + 3^{n} + \sum\limits_{i=1}^{n}3^{n-i}2^{\left(\sum\limits_{k=1}^{i}w_k\right)}}{2^{\left(\sum\limits_{j=1}^{n+1} w_i\right)}}$$

Answer 1

There's a more standard way of expressing this, which is to compile the sequence of divisions by two into a strictly increasing list of powers of two, and it comes up a bit clearer. Then you get on the left hand side of your equation a number whose power of two factor is bigger than that of all the terms on the right.

It's easier to see if you let the powers of $2$ accumulate by using the function $C(x)=3x+2^{\nu_2(x)}$. Observe that this function commutes with multiplication by $2$ so $C(2x)=2C(x)$. This means you can keep applying it directly to even numbers too, no need to find the odd number at each composition.

You can see the pattern emerge by manually composing:

$C^3(x)=3(3(3x+2^{k_0})+2^{k_1})+2^{k_2}=3^3x+3^2\cdot2^{k_0}+3\cdot2^{k_1}+2^{k_2}$

Then if you want you can divide by whatever the final power of two is, to yield a 5-rough integer.

If you compose up to $n=$infinitely many times you get a sequence that converges to $0$ in the 2-adic metric space independently of whether the Collatz conjecture is true or not:

$$C^n(x)=3^nx+\sum_{i=0}^n 3^i\cdot2^{k_i}$$ where $2^{k_i}$ is a strictly increasing sequence of powers of $2$. If I understood your question correctly then this is the greatly-simplified version of your formula.

While $C^{n\to\infty}$ converges to $0$ in the 2-adic space, the Collatz conjecture states that for all positive odd numbers $p$ there is some dyadic fraction $x=\dfrac p{2^r}$ such that for some $n$, $C^n(x)=1$. In fact there are infinitely many such dyadic fractions - if $2^r$ gives a solution $n=n_0$ compositions then every element of the sequence $2^{r+2s}:s\in\Bbb N$ gives a solution too, namely $n=n_0+s$.