Structural and logical proof of Collatz conjecture

Finally mystery finished Every number in collatz map has different set of predecessors, and no 2 numbers share same predecessors. This argument is enough to prove collatz conjecter to be true and no cycle will ever exsist. Any cycle to exsist we must have a number that can be its own predecessor while except 1 no number can fullfil this condition. Reverse function of collatz will be like for any number n. (n.2^m -1 )/3 for any n whenever this equation wil give integer answer that number will be valid predecessor for that n. And we can have infinite many predecessors for any n. This equation could only be equal to n when n<3 for any n≥3 this equation couldn't be equal to n itself. And for n the output (valid predecessors) will be different from any other n. So when no number share same predecessors, and no number can be its own predecessor how cycles can form? This ensures every Collatz sequence will have different from each other numbers inside.

Publication Details

Journal
Zenodo (CERN European Organization for Nuclear Research)
Published
2025-01-01
DOI
https://doi.org/10.5281/zenodo.16889448
Primary Topic
Diverse Scientific and Economic Studies
Type
article
Field-Weighted Citation Impact
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article

Structural and logical proof of Collatz conjecture

Zenodo (CERN European Organization for Nuclear Research)
Diverse Scientific and Economic Studies
article

Structural and logical proof of Collatz conjecture

article en

Abstract

Finally mystery finished Every number in collatz map has different set of predecessors, and no 2 numbers share same predecessors. This argument is enough to prove collatz conjecter to be true and no cycle will ever exsist. Any cycle to exsist we must have a number that can be its own predecessor while except 1 no number can fullfil this condition. Reverse function of collatz will be like for any number n. (n.2^m -1 )/3 for any n whenever this equation wil give integer answer that number will be valid predecessor for that n. And we can have infinite many predecessors for any n. This equation could only be equal to n when n<3 for any n≥3 this equation couldn't be equal to n itself. And for n the output (valid predecessors) will be different from any other n. So when no number share same predecessors, and no number can be its own predecessor how cycles can form? This ensures every Collatz sequence will have different from each other numbers inside.

Zenodo (CERN European Organization for Nuclear Research)
Openalex Percentile: Top 37%
Diverse Scientific and Economic Studies
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