Collatz conjecture

This very first article¹ is interesting.

The following formula is obtained by applying $n$ times the reverse function $f$ to a number $a \in \mathbb{N}$:

$$f^{-n}(a) = \frac{1}{3^n} \left( a 2^{\sum\limits_{i=0}^n p_i} - \sum\limits_{i=1}^n 3^{i-1} 2^{\sum\limits_{j=i+1}^n} p_j \right)$$

The problem

$$f(n) = \begin{cases} n/2 & \text{if }n \equiv 0 \mod 2 \\ 3n+1 & \text{if }n \equiv 1 \mod 2 \end{cases}$$

The conjecture

This process will eventually reach the number 1, regardless of which positive integer is chosen initially.

Generator

Any function of the form:

$$\begin{aligned} \mathcal{G} :\ & \mathbb{N} \rightarrow \mathbb{N} \\ & a \rightarrow a \times 2^p ,p \in \mathbb{N} \end{aligned}$$

is defined as a generator for the Syracuse problem because any number leading to a function of this form will fall down to $1$ after $n$ iterations.

Anytime a number $a$ is having the following property, the conjecture is demonstrated:

$$\exists p \in \mathbb{N}, f(a) = 3a + 1 = \mathcal{G}(p)$$

This means that any number respecting the following equation is validating the conjecture:

$$a = \frac{\mathcal{G}(p) - 1}{3} = \frac{2^p - 1}{3}$$

The question is then: is there for any number $a \in \mathbb{N}$, a fixed number of iteration for $f$ leading to a generator function?

If this is the case, the conjecture is demonstrated.

Relation between generators

A generator has several children but a single parent.

Is the following relation truth?

$$\forall b \in \mathbb{N}, \exists! a \in \mathbb{N} / f(\mathcal{G}_b) = \mathcal{G}_a$$

Demonstration

Written, must be copied here

Conclusion

A generator has several children by the application of the function $f$ but has a single parent by the application of the same function.

Neutral element

The generator $\mathcal{G}_1$ is a neutral element for the function $f$:

$$f(\mathcal{G}_1) = \mathcal{G}_1$$

with

$$\mathcal{G}_1 = 2^p, p \in \mathbb{N}$$

Questions:

• Do we have other neutral elements for the function $f$?
• Is there a relation between neutral elements and cycles for the function $f$?

Demonstration

$$\begin{aligned} f(\mathcal{G}_a) & = & \mathcal{G}_a & \in \mathbb{N}\\ 3 \times a 2^p + 1 & = & a 2^q & => q > p\\ a \left( 2^q - 3 \times 2^p \right) & = & 1\\ a 2^p \left( 2^{q-p} - 3 \right) & = & 1\\ a & = & \frac{1}{2^p \left( 2^{q-p} - 3 \right)}\\ a, p, n \in \mathbb{N} & => & a = 1 & \end{aligned}$$

Children of a generator

A generator $\mathcal{G}_a$ is having the following children by application of the reverse function of $f$:

$$\begin{aligned} f^{-1}(\mathcal{G}_a) & = & \frac{a 2^p - 1}{3} \in \mathbb{N}, p \in \mathbb{N} \\ \end{aligned}$$

What are the values of $p$?

All odd numbers can be defined with the following form:

$$a = 3 \left( n \pm 0,2 \right), n \in \mathbb{N^*}$$

Leading to

$$1, 3, 5 - 7, 9, 11 - 13, 15, 17 - 19, 21, 23 - 25, 27, 29 - ...$$

Thus, what are accepted values of $p$ with the following form?

$$\begin{aligned} f^{-1}(\mathcal{G}_a) & = & \frac{a 2^p - 1}{3} & \in \mathbb{N}, p \in \mathbb{N} \\ & = & \frac{3 \left( n \pm 0,2 \right) 2^p - 1}{3} & \in\mathbb{N}, \forall n \in \mathbb{N} \end{aligned}$$

For $a = 3n, n \in \mathbb{N}$

$$\begin{aligned} f^{-1}(\mathcal{G}_a) & = & \frac{a 2^p - 1}{3} & \in \mathbb{N}, p \in \mathbb{N} \\ & = & \frac{3 n 2^p - 1}{3} & \in \mathbb{N}, \forall n \in \mathbb{N} \\ & = & n2^p - \frac{1}{3} & \notin \mathbb{N}, \forall n, p \in \mathbb{N} \end{aligned}$$

Conclusion A generator of type $\mathcal{G}_{3k}, k \in \mathbb{N}$ does not have any children generator.

We can thus simplify the problem. What are values of $p$ for $a = 3 \left( n \pm 2 \right), n \in \mathbb{N}$ leading to:

$$\frac{a 2^p - 1}{3} \in \mathbb{N}$$

For $a = 3n - 2, n \in \mathbb{N}$

$$\begin{aligned} f^{-1}(\mathcal{G}_a) & = & \frac{a 2^p - 1}{3} & \in \mathbb{N}, p \in \mathbb{N} \\ & = & \frac{3 n 2^p - 2^{p+1} - 1}{3} \\ & = & n2^p - \frac{2^{p+1} + 1}{3} & \in \mathbb{N} \text{ if } p+1 \equiv 2k + 1, k \in \mathbb{N} \\ \end{aligned}$$

Conclusion For $a = 3n - 2, n \in \mathbb{N}$, only values of $p$ of the form $p = 2k, k \in \mathbb{N}$ are defining valid children generators for$\mathcal{G}_a$.

For $a = 3n + 2, n \in \mathbb{N}$

$$\begin{aligned} f^{-1}(\mathcal{G}_a) & = & \frac{a 2^p - 1}{3} & \in \mathbb{N}, p \in \mathbb{N} \\ & = & \frac{3 n 2^p + 2^{p+1} - 1}{3} \\ & = & n2^p + \frac{2^{p+1} - 1}{3} & \in \mathbb{N} \text{ if } p+1 \equiv 2k, k \in \mathbb{N} \\ \end{aligned}$$

Conclusion For $a = 3n + 2, n \in \mathbb{N}$, only values of $p$ of the form $p = 2k + 1, k \in \mathbb{N}$ are defining valid children generators for $\mathcal{G}_a$.

Based on previous results, we can build the following array

$n$	$2^n$	$1$	$5$	$7$	$11$	$13$	$17$	$19$	$23$	$25$	$29$	$31$
$0$	$1$	$0$		$2$		$4$		$6$		$8$		$10$
$1$	$2$		$3$		$7$		$11$		$15$		$19$
$2$	$4$	$1$		$9$		$17$		$25$		$33$		$41$
$3$	$8$		$13$		$29$		$45$		$61$		$...$
$4$	$16$	$5$		$37$		$69$		$101$		$...$		$...$
$5$	$32$		$53$		$117$		$181$		$...$
$6$	$64$	$21$		$149$		$277$		$405$		$...$
$7$	$128$		$213$		$469$		$725$		$...$
$8$	$256$	$85$		$597$		$1109$		$1621$		$...$
$9$	$512$		$853$		$1877$		$2901$		$...$
$10$	$1024$	$341$		$2389$		$4437$		$6485$		$...$
$11$	$2048$		$...$		$...$		$...$		$...$
$12$	$4096$	$1365$		$...$		$...$		$...$

Is there a relation between generators values for consecutive valid odd numbers for a given $n$ value?

$$\begin{aligned} \delta^n_\leftrightarrow & = & \frac{\left( b + 6 \right) 2^n - 1}{3} - \frac{b 2^n - 1}{3} \\ & = & \frac{6 \times 2^n}{3} \\ & = & 2 \times 2^2 \\ \delta^n_\leftrightarrow & = & 2^{n+1} \end{aligned}$$

Conclusion In the array given above, the difference between 2 values on the same row is $2^{n+1}, n \in \mathbb{N}$.

Is there a relation between generators values for consecutive values power of $2$ for a given $a$ value?

$$\begin{aligned} \delta^n_{a, \updownarrow} & = & \frac{a 2^{n+2} - 1}{3} - \frac{a 2^n - 1}{3} \\ & = & \frac{a 2^n \left( 2^2 - 1 \right) - 1}{3} \\ & = & \frac{3 a 2^n}{3} \\ \delta^n_{a, \updownarrow} & = & a 2^n \end{aligned}$$

$\mathbb{N}$ mapped

Do we have an equivalence between each $n \in \mathbb{N}$?

If we are observiing the arra accurately, we can extract that if any number of $\mathbb{N}$ can be expressed with:

$$n = \frac{2^{2p} \left( 1 + 6q \right) - 1}{3} \text{ where } p, q \in \mathbb{N}$$

or

$$n = \frac{2^{2p + 1} \left( 5 + 6q \right) - 1}{3} \text{ where } p, q \in \mathbb{N}$$

Order relation between generators

The following order between generators can be easily extracted from the array:

$$1, 5, 13, 17, 11, 7, 37, 49, 65, 43, 229, 305, 203, 541, 721, 961, 5125, 6833, 4555, 6073, 32389, ...$$

Understanding

$$\mathcal{G}_1 \supset \mathcal{G}_5 \supset \mathcal{G}_{13} \supset ...$$

or

$$\begin{aligned} f(\mathcal{G}_5) = \mathcal{G}_1 \\ f^2(\mathcal{G}_{13}) = \mathcal{G}_1 \\ f^3(\mathcal{G}_{17} = \mathcal{G}_1 \\ ... \end{aligned}$$

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