Collatz Sequence Proof (1st Way)

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