Revealing a Binary Pattern Validates 3n+1 Problem for All Positive Integers

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Revealing a Binary Pattern Validates 3n+1 Problem for All Positive Integers

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Budee U Zaman^*

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Fuzzy Number, Fuzzy Difference, Fuzzy Differential: Theory and Applications

Submitted:

15 April 2024

Posted:

15 April 2024

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Abstract

This study delves into a unique binary pattern found within the wellknown 3n+1 problem, or Collatz conjecture. Through careful analysis of the steps in the 3n+1 sequence, we have discovered a special binary representation that captures the behavior of all positive integers undergoing this transformation. With this new understanding, we have provided a solid proof confirming the validity of the 3n+1 problem for all positive integers. Our method goes beyond the need for extensive computational confirmation, providing a simple and elegant resolution to a long-standing mathematical mystery.

Keywords:

Subject: Computer Science and Mathematics - Algebra and Number Theory

1. Introduction

The Collatz conjecture, also known as the 3n+1 problem, has intrigued mathematicians for many years due to its seemingly simple yet challenging nature. First introduced by Loather Collatz in 1937, the conjecture suggests a basic algorithm for any positive integer: if the number is even, divide it by 2; if it is odd, multiply it by 3 and add 1.

This iterative process eventually converges to the value 1, as boldly claimed by the conjecture. Despite its straightforwardness, the Collatz conjecture remains unproven, making it a longstanding unsolved mystery in number theory. Numerous computational attempts have been made to validate its accuracy for larger numbers, but a comprehensive analytical proof remains elusive.

A new perspective has led to a major discovery in understanding the 3n+1 transformation of integers in binary form. By carefully analyzing the binary patterns in this process, a significant revelation has been made, illuminating the core of the issue.

This research explores a unique viewpoint on the Collatz conjecture. Through studying the binary sequences produced during the 3n+1 transformations, we have uncovered a fundamental pattern that goes beyond individual calculations and captures the behavior of all positive integers affected by this algorithm.

Note that each positive odd integer n, definable as

n = \sum_{i = 0}^{x} 4^{i}

, for each

x \in Z^{+}

, needs to be reduced to one by taking one

3 n + 1

step, followed by

2 (x + 1)

successive

\frac{n}{2}

steps.

The

3 n + 1

step that uses an integer in base 2 will demonstrate the veracity of this claim. [1,2,3]

2. Example One

Let

n = \sum_{x = 0}^{n} {(2)}^{2 i} = 21 = 10101_{2}

, then

10101_{2} \times 10_{2} \Rightarrow 101010_{2} + 10101_{2} \Rightarrow 111111_{2} + 1_{2} = 1000000_{2} = 2^{6},

and

\frac{1000000_{2}}{10_{2}} \Rightarrow \frac{100000_{2}}{10_{2}} \Rightarrow \frac{10000_{2}}{10_{2}} \Rightarrow \frac{1000_{2}}{10_{2}} \Rightarrow \frac{100_{2}}{10_{2}} \Rightarrow \frac{10_{2}}{10_{2}} = 1 .

Consequently, compared to their base 10 representation, the base 2 representation of positive integers provides further understanding of the

3 n + 1

problem.

Proof.

Let

O^{+}

be the set of positive odd integers, then

O^{+} = {x \in Z | x = 2 y + 1, y \geq 0, y \in Z} .

□

3. Theorem One

P will stand for the 3n+1 problem. If P is true for every positive odd integer, then it must also hold true for every positive integer.

\forall a \in O^{+} : P (a) \Rightarrow \forall b \in Z^{+} : P (b)

Proof.

First Case:

Let

x \in Z^{+}

, let

n = 2^{x}

. In order to reduce n to 1, x successive

\frac{n}{2}

steps are needed.

Second Case : Multiplication of an odd integer by a power of two

With

n \in O^{+}

and

x \in Z^{+}

, let

y = 2^{x} \cdot n

. In order to get

y = n

, then x consecutive

\frac{n}{2}

steps are needed.

If we consider all positive integers a, the

3 n + 1

problem encompasses every possible transformation that a positive integer can undergo through iterations. Each step either applies the operation

3 n + 1

or removes a factor of 2 through the

\frac{n}{2}

step. Ultimately, this process converges for every integer n to a power of 2, denoted as

2^{x}

, where x is a non-negative integer.

However, the transformation from any arbitrary integer n to

2^{x}

might not be immediately clear due to the interplay between the

3 n + 1

and

\frac{n}{2}

steps. To elucidate this process, we can focus solely on the

3 n + 1

step while compensating for the omission of the

\frac{n}{2}

step. By adjusting the

3 n + 1

operation appropriately, we can still achieve the convergence to

2^{x}

for every positive integer, making the iterative nature of the transformation more apparent. □

4. Example Two

Let

n = 9 = 1001_{2}

, then

3 n + 2^{x}

produces this pattern:

\begin{matrix} 1001_{2} \times 11_{2} \Rightarrow 11011_{2} + 1_{2} = 11100_{2} \\ 11100_{2} \times 11_{2} \Rightarrow 1010100_{2} + 100_{2} = 101100_{2} \\ 101100_{2} \times 11_{2} \Rightarrow 100001000_{2} + 1000_{2} = 100010000_{2} \\ 100010000_{2} \times 11_{2} \Rightarrow 1100110000_{2} + 10000_{2} = 1101000000_{2} \\ 1101000000_{2} \times 11_{2} \Rightarrow 100111000000_{2} + 1000000_{2} = 101000000000_{2} \\ 101000000000_{2} \times 11_{2} \Rightarrow 1111000000000_{2} + 1000000000_{2} = 10000000000000_{2} = 2^{13} . \end{matrix}

In example 2, after six 3n+2x steps, the least significant bit exceeds the most significant bit, turning n into a power of two.

Definition

The least significant bit of s

\in Z^{+}, then LSB = {2^{r} | r \geq 0, r \in Z such that 2^{r} = \frac{s}{t}, t \in O^{+}} .

The least significant bit=LSB

4.1. Theorem Two

The

3 n + 1

step is isomorphic to the

3 n + L S B

step.

Proof.

Let

n_{0} \in O^{+} . Let n_{1} = 3 n_{0} + 1 and n_{2} = \frac{n_{1}}{LSB}, then \frac{3 n_{1} + LSB}{3 n_{2} + 1} = \frac{3 n_{1} + LSB}{3 (\frac{n_{1}}{LSB}) + 1} = L S B

Given the congruence

3 n + LSB \equiv 0 (mod 3 n + 1)

, we can establish isomorphism between the

3 n + LSB

step and the

3 n + 1

step.

Two functions make up the pattern in Example 2. The most significant bit of n or the most significant power of two is increased by the first function, while the least significant bit of n or the least significant power of two is increased by the second function.

Let

m (x)

be the function for repeated multiplication of n by 3 in terms of x, where

x \in Z^{+}

. Then

m (x) = 3^{x + δ} n

Let

lsb (x)

be the function for repeated multiplication by 4 (

3 (LSB) + LSB

) of the least significant bit of n in terms of x, where

x \in Z^{+}

. Then

lsb (x) = 2^{2 (x + δ)}

. □

5. Definition Two

Let f(x) be the function, in terms of x,

x \in Z^{+}

, for the

3 n + LSB

step for

n \in O^{+}

.Then

f (x) = m (x) + LSB (x) = 3^{(x + δ)} n + 2^{2 (x + δ)} .

Let Tlsb(x) be the function that, for every

n \in O^{+}

, returns the true position of the least significant bit of the

3 n + LSB

step in terms of

x \in Z^{+}

. Next

δ = \sum_{x \in Z^{+}} (T lsb (x) - lsb (x))

Example Three

Assume that multiplying n_{k} by 3 produces \dots 001111100 \dots

somewhere in the binary representation of the result; and that the rightmost 1 is LSB = 2^{x} .

Let lsb (x) = T_{lsb} (x) . Adding LSB to n_{k} yields \dots 010000000 \dots

δ = \sum_{x}^{x} Tlsb (x) - lsb (x)

δ = \sum_{x}^{x} (2^{x + 5} - 2^{x + 2})

δ = \sum_{x}^{x} (x + 5 - x - 2)

δ = \sum_{x}^{x} (3) = 3

Example Four

T l s b (x) \leq l s b (x)

Assume that the binary representation of the result, after multiplying

n_{k}

by 3 and adding LSB, is

\dots 001111100 \dots

, and that the rightmost 1 is

L S B = 2^{x}

. Assume

T L s b (x) = L s b (x)

. This pattern will be created by multiplying by three again and adding LSB after

\dots 001111100 \dots t i m e s 3 p l u s 2^{x}

\dots 101111000 \dots t i m e s 3 p l u s 2^{x + 1}

\dots 001110000 \dots t i m e s 3 p l u s 2^{x + 2}

\dots 101100000 \dots t i m e s 3 p l u s 2^{x + 3}

\dots 001000000 \dots

,than

δ = \sum_{x}^{x + 3} Tlsb (x) - lsb (x)

δ = \sum_{x}^{x + 3} (2^{x + 1} - 2^{x + 2})

δ = \sum_{x}^{x + 3} (x + 1 - x - 2)

δ = \sum_{x}^{x + 3} (- 1) = - 4

Given:

δ < 0 \lor δ = 0 \lor δ > 0

If x is assumed to be

x \in Z^{+}

, then

m (x) < l s b (x)

indicates that a power of two is greater than the sum of its powers.

Using Example 2 as an illustration:

m (x) - l s b (x) = 9 \cdot 3^{(x + 2)} - 4^{(x + 2)} = 0 for x \approx 5.6377 .

The integer after the root necessitates that

m (x) < l s b (x)

. In other words, it requires six

3^{n} + LSB

steps for 9 to converge to

2^{13}

5.1. Theorem Three

There is a positive integer x such that

m (x) < l s b (x)

for all positive odd integers n.

For every

n \in O^{+}

\exists x i n Z^{+} (m (x) < l s b (x))

Proof. Case one

Given:

δ \leq - 1

δ \in Z

Assume

n \in O^{+}

and let

m (x) - lsb (x) = 3^{x - δ} n - 4^{x - δ} = 0

x = \frac{l o g (1 / n)}{l o g (3 / 4)} + δ .

Therefore, there exists a unique

x \in R^{+}

such that

3^{x - δ} n - 4^{x - δ} = 0

and

\exists \Rightarrow x \in Z^{+}

such that

(m (x) < lsb (x))

Case Two

Given:

δ = 0

Assume

n \in O^{+}

and let

m (x) - lsb (x) = 3^{x} n - 4^{x} = 0

x = \frac{l o g (1 / n)}{l o g (3 / 4)} .

Therefore, there exists a unique

x \in R^{+}

such that

3^{x} n - 4^{x} = 0

and

\exists \Rightarrow x \in Z^{+}

such that

(m (x) < lsb (x))

Case Three

Given:

δ \geq 1

δ \in Z

Assume

n \in O^{+}

and let

m (x) - lsb (x) = 3^{x + δ} n - 4^{x + δ} = 0

x = \frac{l o g (1 / n)}{l o g (3 / 4)} - δ .

Therefore, there exists a unique

x \in R^{+}

such that

3^{x + δ} n - 4^{x + δ} = 0

and

\exists \Rightarrow x \in Z^{+}

such that

(m (x) < lsb (x))

Since these examples are all-inclusive, it demonstrates that

For every

n \in O^{+}

\exists x i n Z^{+} (m (x) < l s b (x))

For all

n \in O^{+}

, there exists an

x \in Z^{+}

such that

m (x) < lsb (x)

(Theorem 3), therefore

f (x)

converges to

2^{y}

y \in Z^{+}

. And since the

3 n + LSB

step and the

3 n + 1

step are isomorphic (Theorem 2), it can be concluded that if

a_{0} = n

n \in O^{+}

, then...

a_{i + 1} = {\begin{matrix} a_{i} / 2 & f o r e v e n a_{i} \\ 3 a_{i} + 1 & f o r o d d a i \end{matrix}

converges to 1.

Theorem 1 states that the truth applies to all positive integers since the

3 n + 1

issue holds true for all positive odd numbers. As n

\in Z^{+}

, if

a_{0} = n

, then

a_{i + 1} = {\begin{matrix} a_{i} / 2 & f o r e v e n a_{i} \\ 3 a_{i} + 1 & f o r o d d a i \end{matrix}

converges to 1. □

6. Conclusion

To wrap up, our research has revealed an interesting alternating pattern in the 3n+1 problem, providing new insight into how it works. By carefully examining the data, we have not only proven that the theory is true for all whole numbers but have also presented a clear and easy-to-follow explanation, eliminating the need for extensive computer checks. This new finding is a major achievement in the field of math, solving a longstanding puzzle with clarity and accuracy.

References

Budee U Zaman. Collatz conjecture proof for special integer subsets and a unified criterion for twin prime identification. 2023. [CrossRef]
Budee U Zaman. Exploring the collatz conjecture through directed graphs. 2024. [CrossRef]
Budee U Zaman. Validating collatz conjecture through binary representation and probabilistic path analysis. 2024. [CrossRef]

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Revealing a Binary Pattern Validates 3n+1 Problem for All Positive Integers

Abstract

1. Introduction

2. Example One

3. Theorem One

4. Example Two

4.1. Theorem Two

5. Definition Two

5.1. Theorem Three

6. Conclusion

References

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