Routes

Let `obrace(n)` be the route of `n`, `obrace(n)^k` its restriction to the `k` first steps, `hat T_([k])(n)` the corresponding trajectory and `tau_(n,k)` the corresponding thread.

Properties

Sign

The route of `n` takes values on positive or negative integers depending only of the sign of `n`.
`{(AA n>0", " AA i in NN ", " F^((i))(n)>0),(AA n<0 ", " AA i in NN ", " F^((i))(n)<0):}`

Demonstration:
By induction on `i`,

1. At level 0, the proposition is true.
`{(AA n>0", " F^((0))(n)=n>0),(AA n<0 ", " F^((0))(n)=n<0):}`

2. Then let us suppose that the proposition is true up to level `k`:
`{(AA n>0", " AA i<=k ", " F^((i))(n)>0),(AA n<0 ", " AA i<=k ", " F^((i))(n)<0):}`

In particular:
`{(AA n>0", " n_k=F^((k))(n)>0),(AA n<0 ", " n_k=F^((k))(n)<0):}`

The next step can be either `U` or `D`

(a) Case D
`n_(k+1)=n_k/2`
`{(n_k>0 rArr n_(k+1)>0),(n_k<0 rArr n_(k+1)<0):}`

(b) Case U
`n_(k+1)=(3*n_k+1)/2`
`{(n_k>0 rArr n_(k+1)>1/2>0),(n_k<0 rArr n_(k+1)<1/2):}`
and,
`n_k in I_U=[dot(2)]_3`
`n_(k+1)<1/2 rArr n_(k+1)<=-1<0`

In each case,
`{(n_k>0 rArr n_(k+1)>0),(n_k<0 rArr n_(k+1)<0):}`
which concludes the demonstration. `square`

Repartition in modulo 2 and modulo 3 equivalence classes

Let us note `rho_(i,k)(n)` the proportion of values of `obrace(n)` which are congruent to `i` modulo `k`

Repartition in modulo 2 equivalence classes

Obviously the repartition of Odd and Even Steps within the trajectory is:
`rho_(0,2)(n)=(e(n))/(sigma_(oo)(n))=1-c(n)`
`rho_(1,2)(n)=(o(n))/(sigma_(oo)(n))=c(n)`

Repartition in modulo 3 equivalence classes

To evaluate the repartition of the `n_k` amongst the congruence classes modulo 3, let us look at the forward steps for congruence classes modulo 6:
`{:(6*k mapsto 3*k,,[dot(0)]_6 rarr [dot(0)]_3), (6*k+1 mapsto 9*k+2,,[dot(1)]_6 rarr [dot(2)]_9 sub [dot(2)]_3), (6*k+2 mapsto 3*k+1,,[dot(2)]_6 rarr [dot(1)]_3), (6*k+3 mapsto 9*k+5,,[dot(3)]_6 rarr [dot(5)]_9 sub [dot(2)]_3), (6*k+4 mapsto 3*k+2,,[dot(4)]_6 rarr [dot(2)]_3), (6*k+5 mapsto 9*k+8,,[dot(5)]_6 rarr [dot(8)]_9 sub [dot(2)]_3):}`

The proportions verify:
`{(rho_(0,3)=rho_(0,6)),(rho_(1,3)=rho_(2,6)),(rho_(2,3)=rho_(1,6)+rho_(3,6)+rho_(4,6)+rho_(5,6)):}`
and:
`{(rho_(0,6)=rho_(0,3)*rho_(0,2)=rho_(0,3)*(1-c)), (rho_(1,6)=rho_(1,3)*rho_(1,2)=rho_(1,3)*c), (rho_(2,6)=rho_(2,3)*rho_(0,2)=rho_(2,3)*(1-c)), (rho_(3,6)=rho_(0,3)*rho_(1,2)=rho_(0,3)*c), (rho_(4,6)=rho_(1,3)*rho_(0,2)=rho_(1,3)*(1-c)), (rho_(5,6)=rho_(2,3)*rho_(1,2)=rho_(2,3)*c):}`

Solving the system we have:
`{(rho_(0,3)=0),(rho_(1,3)=1/(2-c)),(rho_(2,3)=(1-c)/(2-c)):}`

Of course, this repartition is valid for infinite (or very long) trajectories. For real (or smaller) trajectories we must take into account that they can begin with one or several numbers in `[dot(0)]_3`

The first number as a probability `1/3` of to belong to `[dot(0)]_3`, each successive number a probability of `1/2` to remain in `[dot(0)]_3`.

The average quantity of numbers in a route is then:
`m_(0,3)=1/3*sum_(i=0)^oo(1/2^i)=2/3`

The probability to be in `[dot(0)]_3` is then `m_(0,3)/sigma_(oo)`

In another hand, the height, the Total Stopping Time and the completeness verify:
`sigma_(oo)=u/(ln(2)-c*ln(3))`

The probability to be in `[dot(0)]_3` is:
`rho_(0,3)=(2*(ln(2)-c*ln(3)))/(3*u)`

The repartition for a given height and a given completeness is then:
`{(rho_(0,3)=(2*(ln(2)-c*ln(3)))/(3*u)),(rho_(1,3)=(1-rho_(0,3))/(2-c)),(rho_(2,3)=((1-rho_(0,3))*(1-c))/(2-c)):}`

For example, the real repartition for trajectories of numbers between 10000 and 20000 and the model are shown on the figure bellow: