Collatz Sequence Proof Easy Way

22 May 2026, Version 1
This content is an early or alternative research output and has not been peer-reviewed by Cambridge University Press at the time of posting.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

The document proposes an attempted proof of the Collatz Conjecture by asserting that every Collatz sequence ( S(n) ) eventually reaches the loop ({4,2,1}). The author introduces “Taha’s Collatz Fact 1 (TCF1),” which claims that for any starting value ( n \in \mathbb{N}^+ ), the iterative sequence ( S(n) ) shares the same invariant loop ( IS(n) = {4,2,1} ). The text illustrates this with explicit sequences for ( n = 1 ) through ( n = 7 ), each ending in the standard Collatz cycle. The argument then attempts an induction-based extension: assuming ( IS(r) = {4,2,1} ) for all values up to ( r ), the author analyzes the cases ( r+2 ) even and ( r+2 ) odd, concluding that both lead to the same invariant loop. The document ends with the claim that the induction covers all natural numbers, implying ( IS(n) = {4,2,1} ) for all ( n \in \mathbb{N}^+ ). Cited lines: “S(n) = {a, b, c, … , t} = IS(n)” and “IS(n) = {4,2,1} ∀n ∈ (N_even ∪ N_odd) = N+”.

Keywords

• Collatz Conjecture • Collatz Sequence • 3n + 1 Problem • Induction Argument • Invariant Set • Loop {4
2
1} • Even/Odd Case Analysis • Taha’s Collatz Fact (TCF1) • Natural Numbers • Sequence Termination

Comments

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Comment number 1, Peter M: Jul 27, 2026, 05:54

Your induction is incorrect; specifically when you attempt to conclude lS(n)={4,2,1} for all even n. Suppose lS(n)={4,2,1} for all n \in {1,2,3,...,r} where r is even. Then lS(r+2) = {4,2,1} which is correct. Same with r+4, r+6, and so on up to 2r. However, you cannot say the same about 2r+2 because you have not shown lS(n+1) = {4,2,1} follows from the inductive hypothesis. This is precisely where your inductive step fails. This attempt at a proof will be rejected anywhere you might attempt to submit it. You need to do your research instead of attempting to solve these problems blindly. You should also read: https://web.williams.edu/Mathematics/sjmiller/public_html/jntnewpolicy/. This addresses why you should not be attempting these problems unless you are an expert in the literature around them, which you are obviously not.

Response,
Peter M :
Jul 27, 2026, 05:55

*Sorry, I meant lS(r+1), not lS(n+1).