Let Q be a trajectory beginning with x0 :
Q = {xk}, k Î [0,s¥(x1)], xk+1= T(xk)
We note Qi the partial trajectories :
Qi = {xk}, k Î [0,i]
And ti the associated thread :
A trajectory takes values on positive or negative integers only depending of the sign of x0
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The repartition of the xk among the congruence classes modulo 2 is O(x0) odd and E(x0) even :
p0,2 = O(x0)/s¥(x0) = c
p1,2 = E(x0)/s¥(x0) = 1-c
To evaluate the repartition of the xk among the congruence classes modulo 3, let us look at the forward steps for congruence classes modulo 6 :
The probabilities are such as :
With :
Solving the system we have :
Of course, this repartition is for infinite (or very long) trajectories, for real (or smaller) trajectories we musttake into account that they can begin with one or several numbers in [0]3
The first number as a probability of 1/3 to be in [0]3, the next of 1/2 if the first is in [0]3 and so on.
The average quantity of numbers in [0]3 in a trajectory is then :
The probability to be in [0]3 is then n0,3/s¥ and as the height, the Total Stopping Time and the completeness verify,
The probability to be in [0]3 is :
The repartition for a given height and a given completeness is then :
For example, the trajectories of numbers between 10000 and 20000 gives the real repartitions and the model as follows :
Figure 1 : Equivalence Classes modulo 3
among trajectories of given completeness c