Stopping Times of Syracuse Integer Sequences on ℤ/2^n ℤ
2021
The Collatz Conjecture affirm that for all positive integer m, the integer sequence defined by T(m) = m/2 if m is even and T(m) = (3m+1)/2 else, there is an integer k such as T^k (m)=1. It has been proved that another equivalent formulation of the conjecture is to say that all trajectories of a Collatz integer sequence has a finite stopping time, which means that all trajectories finish to be lower than the starting term of the integer sequence. This consideration is obvious for all even number m and can be easily proved for all integer m≡1[4]. But the question becomes extremely complex if we want to analyze the behavior of integers m≡3[4]. I. Krasikov and J. C. Lagarias [1]2002 proved in 2002 that the number of integers below x, ԥ1(x), which reach 1 under iteration satisfy the inequality ԥ1(x)>x^0,84. In the present work, we have based our approach on the second formulation of the conjecture which affirm that all trajectories has a finite stopping time. We are going to analyze the behavior of all Syracuse integer sequences starting from odd integers belonging to ℤ/2^n ℤ for successive integer values of n and we'll state that the percentage of integers P(n)=S(n)/2^n tends toward 0 when n→∞, where S(n)= #{a∈ ℤ/2^n ℤ, σ(a)>n} and σ(a) is the stopping time of a Syracuse sequence starting from a. We'll start on a computational approach to study the behavior of P(n) for n up to 41 and in a second time thanks to an exact formulation of hn (n) = #{a ∈ ℤ/2^n ℤ , σ(a) = n}, we'll state that P(n) tends toward 0 when n→∞, We'll finish by giving some arguments to conjecture that the maximal stopping time of all Syracuse integer sequences starting from an integer a<2^n, would be below 20n.
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