Collatz Sequence Proof (2nd Way)

10 February 2025, Version 6
This content is an early or alternative research output and has not been peer-reviewed by Cambridge University Press at the time of posting.

Abstract

Abstract of Collatz Sequence Solution (2nd Way) I discovered a fact in Collatz Sequence: S (n) = {a, b, c,…,w} ⇒ LS(n) = LS(a) = LS(b) = LS(c) = …. = LS (w) … (Fact1) n, a, b, c, w, r ∈ N_+, LS(n) = {4,2,1} when n ∈ {1,2,3,4,5, …, r-1} Let LS(r) = {4,2,1} when n ∈ Z = {1,2,3,4,5, …, r-1, r} If (r+2) ∈ (N_even), and I proved LS(r+2) = {4,2,1} ⇒ LS(n) = {4,2,1} for all even numbers. If (n) ∈ (N_odd), and I proved LS(n) = {4,2,1} for all odd numbers. Therefore LS(N) = {4,2,1} for all n ∈N_+

Keywords

Collatz
2nd Way Collatz
Unsolved Math
Taha
Number Theory

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Comment number 2, Taha Muhammad: Sep 01, 2026, 02:40

One person: This proof is completely wrong because the induction is incorrect. If the Collatz function enters the cycle {1, 4, 2} for all numbers in {1, 2, 3, ..., r} where r is an even number, then the third case (iii) only proves that even numbers r+2, r+4, ..., 2r enter the cycle {1, 4, 2}. It does not prove that even numbers greater than 2r enter the cycle {1, 4, 2}, so you cannot use this assumption in the fourth case (iv) when considering the even number 3n+1. My Answer1: Your question is wrong because the term after r is r+2, not 2r. Same Person: You need to do your own research and read peer-reviewed papers on this postulate. Start with the comprehensive paper by Lagarias, as well as Tao's paper, which proves that almost all Collatz sequences eventually become bounded by an increasing function. My Answer 2: I don't need advice. I do not need follow others' steps because I am under control of solving Collatz. Same Person: Both are experts on the Collatz postulate, while you are not. My Answer 3: No need for attack. I have millions of honorable professors are looking at my humble solutions at YouTube, X.com, my books at Amazon.com, my documentary film by Film maker Dr. Aaron Dye, Washington University.

Response,
Peter M :
Sep 01, 2026, 07:10

"Your question is wrong because the term after r is r+2, not 2r." Did you not even read my response? You are claiming that if the conjecture is true for all numbers {1,2,3,...,r} for some even r, then it must be true for all even numbers. This is not even close to justified. Step ii) can only show that the conjecture holds for the numbers r+2, r+4, ..., 2r. Once you get to 2r+2 then your proof fails because you have not shown the conjecture holds for r+1. "I have millions of honorable professors are looking at my humble solutions at YouTube, X.com, my books at Amazon.com, my documentary film by Film maker Dr. Aaron Dye, Washington University." I am calling BS on that. Not a single reputable math professor will look at your solution and think that it's worth anything. You are objectively wasting your own time, and others' time. I really don't care how many people you think looked at your solution or the methods you've tried to communicate your solutions.

Comment number 1, This comment has been removed by the moderator.: Jul 17, 2026, 05:10
This comment has been removed by the moderator.