1. Introduction

2. The Conjecture and Related Conversions

2.1. Collatz Inverse Operation (CIO)

Let $n\in {\mathbb{N}}^{+}$and$a\in {\mathbb{N}}_{odd}$ ; for a to be converted to $2^n$ by the Collatz operation (CO), it must satisfy the following equation,

$$3.a+1=2^n$$

then,

$$a=\frac{2^n-1}{3}$$

(1)

Lemma 2.4. In (1) $a=\frac{2^n-1}{3}$, a cannot be an integer if n is a positive odd integer.

Proof. If n is a positive odd integer, we can take $n=2m+1$ ($m\in \mathbb{N}$), then substituting $2m+1$ for n in (1) we get,

$$a=\frac{2^{2m+1}-1}{3}$$

(2)

if we factor $2^{2m+1}+1$,

$2^{2m+1}+1=(2+1)(2^{2m}-2^{2m-1}+2^{2m-2}-\dots +1)=\mathbf{3}.k$

we get $3k$, which is a multiple of 3 ($k\in {\mathbb{N}}_{odd}$).

Since $2^{2m+1}-1=(2^{2m+1}+1)-2=3.k-2$,

($3k-2$) is not a multiple of 3 ($k\in {\mathbb{N}}_{odd}$), so in (2) a is not an integer for any number n.

If we substitute $2n$ for n in (1), we get equation

$$a=\frac{2^{2n}-1}{3}$$

(3)

Lemma 2.5. In (3) $a=\frac{2^{2n}-1}{3}$, for each number n there is a different positive odd integer a, ($n\in {\mathbb{N}}^{+}$).

Proof. When we factorize $2^{2n}-1$ for all n numbers, ($n\in {\mathbb{N}}^{+}$), if

$n=1,(2^2-1)=(2-1)(2+1)=\mathbf{3}.1$

n=2, (24 -1) = (2-1)(2+1)(22+1)=3.5

n=3, (26 -1) = (23-1)(23+1)=3.3.7

n=4, (28-1) = (2-1)(2+1)(22+1)(24+1)=3.(...)

n=5, (210-1) = (25- 1)(25+1)= (2-1)(24+...+1)(2+1)(24- ...+1)=3.(...)

n=6, (212-1)= (23- 1)(23 + 1)(26+1)=3.3.(...)

n=7, (214 -1)= (27- 1)(2+1)(26- 25+...+1)=3.(...)

...

...

if we substitute 2n to $m^2$,(m ∈${\mathbb{N}}^{+}$);

$(2^{2^m}-1)=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)\dots \dots .(2^{2^{m-1}}+1)=\mathbf{3}.(\dots )$

...

$(2^{2n}-1)=(2^{x_1}-1)(2^{x_1}+1)(2^{x_2}+1)(2^{x_3}+1)\dots .(2^{x_{n-1}}+1)(2^{x_n}+1)$ or

$(2^{2n}-1)=(2^{x_1}-1)(2^{x_1}+1)$ in these equations, $x_1$ is a positive odd integer and $x_2$, $x_3$, $x_4$ ... $x_n$ are positive even integers. Since $x_1$ is a positive odd number,

$(2^{x_1}+1)=(2+1)(2^{x_1-1}-2^{x_1-2}+2^{x_1-3}-\dots +1)=\mathbf{3}.(\dots )$ so,

$(2^{2n}-1)=\mathbf{3}.(\dots )$

Since each of these numbers has a multiplier of 3, we can find positive odd integers a for all n, and when we apply Collatz operations to these a numbers, we always get 1.

$2^{2n}+1$ is not a multiple of 3, since $2^{2n}-1$ is a multiple of 3, for all n ($n\in {\mathbb{N}}^{+}$).

In (3),

$$a=\frac{2^{2n}-1}{3};$$

If $n=1$, $a{}_1=1$

$n=2$, $a{}_2=5$

$n=3$, $a{}_3=21=3.7$

$n=4$, $a{}_4=85$

$n=5$, $a{}_5=341$

... ...

Corollary 2.6. We get a set A with infinite elements, these numbers reach 1 when we apply the Collatz operations. This is because when we apply the Collatz operations to these numbers, they become $2^{2n}$ numbers (Remark 2.3). (In the following sections, we will refer to the elements of the set A and other numbers that satisfy the Collatz conjecture as Collatz numbers).

Example 2.7. 5 → odd number, 16 →8→4→2→1.

21→ odd number, 64→32→16→8→4→2→1.

If we can generalize the elements of the set $A=\{1,5,21,85,341,1365,5461,21845,87381,\dots \}$ to all positive odd numbers, we have proved the Collatz conjecture.

2.3. Conversion of Collatz Numbers to all Positive Odd Integers

In the previous sections, when we applied the Collatz operations, we called the numbers that reached 1 as Collatz numbers. Now let’s see how these Collatz numbers can be converted to all positive integers.

If we apply the Collatz inverse operations [equations (4) or (5)] continuously to each Collatz number, we get infinitely many new Collatz numbers.

${\mathbb{N}}_{odd}$→ Set of A →$2^{2n}$→ 1 (conversion direction of numbers with CO).

${\mathbb{N}}_{odd}$← Set of A ←$2^{2n}$ (conversion direction of numbers with CIO).

Example 2.13. A small fraction of the Collatz numbers, that can be converted to $2^4$, i.e. 1, by applying CO. These numbers are obtained by applying the CIO to each of the numbers. New numbers are obtained by repeatedly applying CIO to any number that is not a multiple of 3.

Example 2.14. Collatz numbers that can be converted to $2^6$, i.e. 1, by applying CO.

21 ←$2^6$ There are no other Collatz numbers. Because the resulting number is a multiple of 3. (Lemma 2.12).

Lemma 2.15. There is only one different Collatz number, which is converted to each of the numbers $2^{6n}$; ($2^6$, $2^{12}$, $2^{18}$, $2^{24}$ ...) $(n\in {\mathbb{N}}^{+})$.

Proof. If we factor $2^{6n}$ - 1,

$2^{6n}-1=(2^{3n}-1)(2^{3n}+1)$, there is always a multiplier of ($2^3+1$) in this equation. Because when we factor ($2^{3n}-1$) and ($2^{3n}+1$);

if n is even, ($2^{3n}-1$) = ... ($2^{3f}+1$), $3f$ is odd integer.

if n is odd in ($2^{3n}+1$), $3n$ is odd integer.

And if $3f$ or $3n$ are odd,

$2^{3n}+1=(2^3+1)(2^{3n-3}-2^{3n-6}+2^{3n-9}-2^{3n-12}+2^{3n-15}\dots +1)$

$2^{3f}+1=(2^3+1)(2^{3f-3}-2^{3f-6}+2^{3f-9}-2^{3f-12}+2^{3f-15}\dots +1)$ Therefore $2^{6n}-1=(2^3+1).(\dots )=9$. (odd integer)

And, when we divide ($2^{6n}-1$) by 3, we get only one Collatz number. We can’t obtain another Collatz number because it is a multiple of 3 (Lemma 2.12).

Example 2.16. There is only one different Collatz number for each of $2^{6n}$, because the resulting Collatz numbers are the multiples of 3 (Lemma 2.12).

$21\leftarrow 2^6$

$1365\leftarrow 2^{12}$

$87381\leftarrow 2^{18}$

$5592405\leftarrow 2^{24}$

... ...

But, for $n\in {\mathbb{N}}^{+}$, $k\in {\mathbb{N}}^{+}$ and $n\ne 3k$, when all $2^{2n}-1$ numbers (except $2^{6n}-1$) are divided by 3, we get positive odd integers that are not multiples of 3. This is because when we factor $2^{2n}-1$, there is only one multiplier of 3.

$(2^{2n}-1)=(2^{x_1}-1)(2^{x_1}+1)(2^{x_2}+1)(2^{x_3}+1)\dots .(2^{x_{n-1}}+1)(2^{x_n}+1)$ or

$(2^{2n}-1)=(2^{x_1}-1)(2^{x_1}+1)$

In these equations, $x_1$ is a positive odd integer and not a multiple of 3, and $x_2$, $x_3$, $x_4$, ... $x_n$ are positive even integers. $(2^{x_1}-1),(2^{x_2}+1),(2^{x_3}+1),\dots (2^{x_{n-1}}+1)$ and $(2^{x_n}+1)$ do not have a multiplier of 3 (Lemma 2.4. and Lemma 2.5). And since $x_1$ is not a multiple of 3, $(2^{x_1}+1)$ has only one multiplier of 3, so an infinite number of Collatz numbers are converted to each of the numbers $2^{2n}$ with CO.

Example 2.17. A small fraction of the Collatz numbers that can be converted to $2^8$, i.e. 1, by applying CO. These numbers are obtained by applying the CIO to each of the numbers. New numbers are obtained by repeatedly applying CIO to any number that is not a multiple of 3.

Similarly, all positive odd integers are converted to $2^{2n}$ ( $n\in {\mathbb{N}}^{+}$ ), i.e. to 1 by applying CO. These numbers are obtained by repeatedly applying the Collatz inverse operations to each element of the set A and the Collatz numbers generated from these numbers.

Lemma 2.18. We obtain new Collatz numbers by applying the Collatz inverse operation ($\frac{2^m.a_n-1}{3}$) ($m\in {\mathbb{N}}^{+}$), to the Collatz numbers, all of these Collatz numbers are different from each other.

Proof. Let $a_1$ and $a_2$ be arbitrary Collatz numbers, when we apply the Collatz inverse operation to each of them, the resulting numbers are $b_1$ and $b_2$. If $b_1=b_2$ then,

$\frac{2^m.a_1-1}{3}=\frac{2^t.a_2-1}{3}$ and $2^m.a_1=2^t.a_2$ for odd positive integers, must be $a_1=a_2$ and $m=t$.

Lemma 2.19. If $a_n\equiv$ 0 (mod 3) and $a_n$ numbers are odd Collatz numbers, we can derive $a_n$ from other Collatz numbers.

Proof. If $a_n\equiv$ 0 (mod 3) and $a_n$ numbers are odd Collatz numbers, then by applying CO to $a_n$ , we find odd positive integers $b_n$. Since $a_n$ numbers are the Collatz numbers, $b_n$ numbers are also the Collatz numbers and $b_n\neg \equiv 0(mod3)$.

$a_n$→$b_n$ (apply CO to $a_n$)

$a_n$←$b_n$ (apply CIO to $b_n$)

since Collatz numbers cover $b_n$ numbers, by applying the Collatz inverse operation to $b_n$ we get $a_n$, $\frac{2^t.b_n-1}{3}=a_n$

Corollary 2.21. By applying the Collatz inverse operations [Equation (3)] to the numbers $2^{2n}$, we get a set A with infinite elements of positive odd numbers. The elements of the set A are the Collatz numbers. We get new Collatz numbers by applying Collatz inverse operations [Equation (4) or (5)] to each element of this set A. From these new infinite Collatz numbers, infinitely many new Collatz numbers are formed by applying the Collatz inverse operations (CIO) again and again, and this goes on endlessly. So we get the whole set of positive odd numbers as Collatz numbers and we prove the Collatz conjecture for all ${\mathbb{N}}^{+}$ (Remark 2.2).

3. The Absence of any Positive Integer other than Collatz Numbers

4. Conclusion

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