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Collatz Sequence

12 July 2024, Version 2
Working Paper
This content is an early or alternative research output and has not been peer-reviewed by Cambridge University Press at the time of posting.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 Solving Collatz Sequence I started from a loop containing all positive numbers and performed Collatz law on each number x in the loop, obtained algebraic equations, and defined the values of x, which not belong to the set of natural numbers except 1. Then I applied Collatz laws and obtained the elements of the loop, which are {4, 2, 1}. S (n)= {n/2 or 3n+1, ..., 4,2,1} for all N+. Thus LS (n) = {4,2,1} for all N+. I added extra work to support my proof: 1- Graph 2- Sketch 3- Chart 4- Table 5- Pattern 5- [If LS (n)= {4,2,1} for r = 3k] True Then I proved that [LS (n) = {4, 2, 1} for r+3= 3(k+1)] True 6- [If LS (n) has no loop if for r = 3k+h, h<3, h is a natural number] True Then I proved that [LS (n) has no loop if for r+1 = 3k+h+1=3k+2, h=1] True Therefore: LS (5) = LS (10^100) = LS ((10^1000) + 7) = LS (47) = ... = LS (1) = {4, 2, 1} only for all n in N+ Note: r = # of elements in a loop. K = 1 or 2

Keywords

Collatz
Sequence
Lothar
Taha
UK
USA
Kurdistan

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Comment number 1, Taha Muhammad: Sep 30, 2024, 20:18

∵ N_+={3k} ∪ {3k+h}, When h=1, or h=2 ∴ loop S (n) = {1,2,4}, ∀n ∈ N_+

Version History

Jul 12, 2024 Version 2

Version Notes

Proof: r+1=3(k)+2 x= (3^k x+3^(k-1) (2^0 )+3^(k-2) (2^2 )+3^(k-3) (2^4 )+⋯+3^(k-k+1) (2^(2(k-2)) ) 〖+2〗^(2(k-1)))/[2^(3k+2-k)] was [2^(3k+1-k)] false because r+1=3(k)+2 x= (3^k x+3^(k-1) (2^0 )+3^(k-2) (2^2 )+3^(k-3) (2^4 )+⋯+3^(k-k+1) (2^(2(k-2)) ) 〖+2〗^(2(k-1)))/2^(2k+2) 〖(2〗^(2k+2)-3^k)x=3^(k-1) (2^0 )+3^(k-2) (2^2 )+3^(k-3) (2^4 )+⋯+3^(k-k+1) (2^(2(k-2)) ) 〖+2〗^(2(k-1)) let k=4 〖(2〗^10-3^4)x=3^3 (2^0 )+3^2 (2^2 )+3^1 (2^4 ) 〖+2〗^6 943x=175⇒x∉N_+

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DOI

10.33774/coe-2024-c4f7p-v2
D O I: 10.33774/coe-2024-c4f7p-v2 [opens in a new tab]

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The author(s) have declared they have no conflict of interest with regard to this content

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