down the road. This paper was developed to numerically specify the. Collatz sequence. Also, the specification was set up to lead towards some kind of proof.
The Collatz Conjecture: Determining an Infinite Convergent Sequence February 9, 2017 Abstract The field of number theory is vital in mathematics. Vitality of this field includes and is not limited to the distribution of prime numbers and cryptography. One of the most important unsolved mathematics problems is the Collatz conjecture. This conjecture consists of a sequence to determine whether it converges to one for all positive initial integer values. The conjecture may hint to other findings later on down the road. This paper was developed to numerically specify the Collatz sequence. Also, the specification was set up to lead towards some kind of proof.
Introduction In 1972, J.H. Conway demonstrated a generalization of the Collatz conjecture to be algorithmically unreachable [1]. Then, thirty-eight years later a man by the name of Jeffrey Lagarias stated this problem to be extraordinarily difficult and completely unsolvable in present day mathematics [2]. My mindset disagrees with this conjecture being impossible. Algorithms are always developed with effort. In this case, I believe it takes more effort, imagination and belief that locating an algorithm is indeed possible. Does the Collatz sequence converge to one for all positive initial integer values? This paper was developed to explain how an algorithm can be obtained.
Body The sequence can begin at the even integers by choice. We can also choose to begin at the odd integers. Let an even number be 2n where n is our initial 1
positive integer. Let an odd initial integer n be tripled and added by one giving us n = 3p + 1. The goal is to develop and algorithm that produces Collatz’s sequence of convergent integers. I will provide examples of the even and odd initial integers and match them with Collatz’s sequence. Before I begin showing a potential sequence, let’s define an arbitrary algorithmic sequence an + b, where a, b, n ∈ Z. I will begin using a couple of integers as the examples. Let the first initial positive even integer be 28. The Collatz sequence is as follows: 28 → 14 → 7 → 22 → 11 → 34 → 17 ↓ 16 ← 5 ← 10 ← 20 ← 40 ← 13 ← 26 ← 52 ↓ 8 → 4 → 2 → 1. Using our initial even number 2n, we can see where the Collatz sequence begins. Since we began with an even positive integer, we half it to obtain the integer we began with. When n is odd, we triple it and add one, correct? Numerically, we can see that, 2n → n → 3n + 1. If an integer n continues to become even, we continue halving it obtaining, 2n → n →
n . 2
I believe the difficulty of this sequence is remaining in the set of integers. Since an arbitrary Collatz sequence was defined in the set of integers, lets define an Even and Odd Collatz sequence. These two sequences will be built from the arbitrary sequence an + b. If b is even and a is odd, multiply a with 2n to get a(2n) + b. The integer a multiplied by 2n will make a even. When a and b are both even, half it obtaining an+b . If a and b are both odd triple 2 it and add one such that 3(an + b) + 1. When a is even and b is odd, does that number exist? Showing this would give us, an + b → (2n)n + (2n + 1) → 2n2 + 2n + 1 or an + b → (2n)n + (3n + 1) → 2n2 + 3n + 1. In order for the algorithm to generate correctly, let the even integer be 2n + b where b is the initial even number 28. So, since 2n + 28 is even, we half it, 2n + 28 → n + 14. 2
Since a is odd and b is even, multiply 1 by 2n to get 2n + 14 then half it, 2n + 14 → n + 7. Notice that b is odd and a is the default odd number 1 on n + 7, we triple (n + 7) and add one such that, n + 7 → 3n + 22 when simplified. Notice that a is odd and b is even again so lets repeat the multiplication process of 2n to a giving us 3(2n) + 22 to continue the sequence, 3n + 22 → 3n + 11. Repeating the sequence from the initial even integer given is illustrated such that, 2n + 28 → n + 14 → n + 7 → 3n + 22 → 3n + 11 → 9n + 34 → 9n + 17 ↓ 81n + 10 ← 81n + 20 ← 81n + 40 ← 27n + 13 ← 27n + 26 ← 27n + 52 ↓ 81n + 5 → 243n + 16 → 243n + 8 → 243n + 4 → 243n + 2 → 243n + 1. Notice that the sequence of 0 b0 is following the Collatz sequence as 0 a0 grows exponentially. Now, let’s focus on when an integer n is odd. Let’s take 17 as an example for our initial integer n. The Collatz sequence of 17 is numerically as follows, 17 → 52 → 26 → 13 → 40 → 20 ↓ .1 ← 2 ← 4 ← 8 ← 16 ← 5 ← 10 There exists an algorithm also for initial odd initial integers as well. This algorithm mixes the even integers with the odd where we begin the Collatz sequence at 3n + 1. Since 17 is odd, we have the initial sequence 3n + b where b equals 17. Then, 3n + 17 → 9n + 52 → 9n + 26 → 9n + 13 → 27n + 40 → 27n + 20 ↓ .81n + 1 ← 81n + 2 ← 81n + 4 ← 81n + 8 ← 81n + 16 ← 27n + 5 ← 27n + 10 Since 17 is odd, letting it be initial number n will give us the second integer in the 3n + 1 sequence. If we want to begin the sequence with 17, let b equal 17 as above. Then, follow the developed algorithm from there. There exists an algorithmic sequence (an + b) converging to 1 for all positive initial n-integers. 3
Theorem 1 (∃ (an+b) ∈ Z converging to 1, ∀ n ∈ Z.). Let a, b, n be positive Z0 s. Then, there exists an algorithmic sequence (an+b) converging to 1 for all positive initial n-integers. Proof. Let a = 3 and b = 1 to obtain 3n+1. (3n+1) represents an entire initial integer. If we want to begin with an initial n-integer of 19, 3(19)+1 = 58, would 58 be our first integer? Actually, beginning with our odd integer of 19 would make us triple it and add one giving us the second integer in the (3n + 1) sequence. An odd integer can also begin with the expression of (2n + 1). As we all know, an even integer begins with the expression of 2n. Going back to beginning with integer 19, we let (3n + 1) = 19 where n = 6. If our initial integer is 6, our second integer in the sequence will be 3 due to it being an even integer and not 19. The exponential growth of integer ’a’ implies the sequence to be infinite for all positive n-integers. Algorithm: If b is even and a is odd, replace n with 2n to get a(2n)+b. This will make a even. When a and b is even in (an + b), half it. If (an + b) is odd where a and b are both odd, triple it and add one such that 3(an + b) + 1.
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References [1] Conway, J.H. (1972). Unpredictable Iterations, Proc. 1972 Number Theory Conf. Univ. of Colorado, Boulder. pp. 49 - 52. [2] Lagarias, J.C., ed. (2010). The Ultimate Challenge: the 3x+1 problem. Providence, R.I.: American Mathematical Society. p. 4.
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