Total Stopping Time

Total Stopping Time Model

Classes

In that model we estimate the number of elements of the Total Stopping Time Classes and Subclasses.

The definition of `F`, can be expressed as a definition of the successors:
`{(2*n prec n),(2*n+1 prec 3*n+2):}`

Predecessor sets are obtained by:
`{({n} succ {2*n}),({3*n+2} succ {2*n+1,6*n+4}):}`

For forward steps (successors) Up: `U` and Down: `D`
`U: {(2NN+1->NN),(n|->(3*n+1)/2):}`
`D: {(2NN->NN),(n|->n/2):}`
For backward steps (predecessors) Top: `T` and Bottom: `B`
`T: {(NN->2NN),(n|->2*n):}`
`B: {(3NN+2->2NN+1),(n|->(2*n-1)/3):}`

Each integer `n` has `2*n` as predecessor by `T` and if `n` is congruent to `2` modulo `3`, it has `(2*n-1)/3` as another predecessor by `B`.
If an integer `n` as a Total Stopping Time `sigma_(oo)(n)`, each predecessor of `n` has a Total Stopping Time `sigma_(oo)(n)+1`

Further if `n` has a number of Odd Steps `o(n)`, its predecessor `2*n` has a number of Odd Steps of `o(n)`, and its predecessor `(2*n-1)/3`, when it exists, has a number of Odd Steps of `o(n)+1`.

If we consider the Class `R_k`, and its subclasses `R_(k,o)`, each element of `R_(k,o)` has a predecessor in the Class `R_(k+1,o)`, and each element of `R_(k,o)` congruent to `2` modulo `3` has a predecessor in the Class `R_(k+1,"o+1")`.

If we define `R_(k,o,p)` as the subset of `R_(k,o)` whose elements are congruent to `p` modulo `3`, `R_(k+1,o)` is the union of the image of `R_(k,o)` by `T` and of the image of `R_(k,o-1,2)` by `B`:
`R_(k+1,o)=T(R_(k,o))uuB(R_(k,o-1,2))`
Further if `R_(k,o-1)` is normally distributed in the congruence classes modulo `3` and each `3^m`, then `R_(k,o)` is normally distributed in the congruence classes modulo `3` and each `3^m`.

Beginning with a given subclass `R_(k,o)` normally distributed in the congruence classes modulo `3` and each `3^m`, we have after `p` steps a binomial distribution of `R_(k+p,"o+i")` classes such that:
`r_(k+p,"o+i")=r_(k,o)*((p),(i))*(1/3^i)`
Where `((p),(i))` is the binomial coefficient `(p!)/(i!*(p-i)!)`, and `1/3` is the basic probability for a number to be congruent to `2` modulo `3`.

It comes that considering the first set `R_0 = R_(0,0) = {1}`, `r_0 = r_(0,0) = 1`,
`r_(k,o)=((k),(o))*1/(3^o)`
`r_k=sum_(o=0)^kr_(k,o)=(4/3)^k`

Finesse

An integer `n` of height `u` has a finesse of:
`gamma(n)=(sigma_oo(n))/u`

if `n` has a number of Odd Steps `o(n)` then the trajectory of `n` counts `o(n)` Odd Steps each increasing the height by `ln(3)-ln(2)` and `sigma_oo(n)-o(n)` Even Steps each decreasing the height by `ln(2)`.

So, the initial height of `n` should be about:
`u~~(sigma_oo(n)-o(n))*ln(2)-o(n)*(ln(3)-ln(2))`
`u~~sigma_oo(n)*ln(2)-o(n)*ln(3)`

And then:
`gamma(n)~~(sigma_oo(n))/(sigma_oo(n)*ln(2)-o(n)*ln(3))`
`gamma(n)~~1/(ln(2)-c(n)*ln(3))`

Maximum Finesse for a given Class

In a given subclass `R_(k,o)`, any element has an approximate height:
`u~~k*ln(2)-o*ln(3)`
an approximate finesse:
`gamma~~1/(ln(2)-k/o*ln(3))`

There is about `((k),(o))*1/(3^o)` elements in that subclass.

The class record of class `R_k` is the smallest element of the non-empty subclass `R_(k,o)` with the highest `o`.

So, we calculate `o_max(k)` such that `r_(k,o_max)>=1` and `r_(k,o_max+1)<1`
`o_max(k)=max_o(((k),(o))*1/(3^o)>=1)`
`gamma_max(k)=k/u_min(k)`
`u_min(k)=k*ln(2)-o_max(k)*ln(3)`

We can build the model which gives the maximum finesse for each class and compare it to the finesse of the current class record-holders.

Comments
We see that the curves for estimated maximum finesse and calculated current maximum finesse have the same shape. But there is a gap between those two curves.
For high classes we can consider that we have not yet found strong enough results, but the gap exist even for the confirmed class records.

Improved Model

The model gives the distribution of predecessors for a given class, with the hypothesis that the elements of that class are normally distributed for congruence modulo `3^m`.
That is obviously not the case for the class `R_0`.

Let us build a model beginning with more consistent classes.
If we begin with the `R_b` class, the number of element of the `R_(k,o)` subclass issued from the `R_(b,i)` subclass is:
`r_(b,i)*((k-b),(o-i))*1/(3^(o-i))`
And the total number of elements of the `R_(k,o)` subclass is:
`r_(k,o)=sum_(i=0)^o(r_(b,i)*((k-b),(o-i))*1/(3^(o-i)))`

We choose `b` such that `r_(k,o)` converges, for example for `k = 1000, o = 600`.
We choose b = 50.

With that optimized model, we now have the curves shown bellow:
And for the first classes, with the confirmed class records (up to 1445 as per september 2021) .

Maximum Finesse and Completeness

Using the binomial model for finesse, we have a model for completeness:
`gamma(n)~~1/(ln(2)-c(n)*ln(3))`
`c(n)~~(ln(2)-1/(gamma(n)))/ln(3)`

A first limit for `c(n)` is given by:
`c(n)<=lim_(gamma(n)->oo)((ln(2)-1/(gamma(n)))/ln(3))=ln(2)/ln(3)`

And, considering the model:
`c_(max)(k)=max_(n in R_(k,o_max(k)))(c(n))=(o_max(k))/k`
`c_(oo)=lim_(k->oo)c_(max)(k)`
`c_(oo)` is solution of the equation:
`c_(oo)*ln(c_(oo))+(1-c_(oo))*ln(1-c_(oo))+c_(oo)*ln(3)=0`
`c_(oo)= 0,6090897679 cdots`

And the upper limit to finesse is:
`gamma_(oo)= 41,67764765 cdots`

We can also estimate the maximum finesse for a number of a given height by:
`gamma_(max)(u)=(k(u))/u`
with `k(u)` such that:
`u=k(u)*ln(2)-o_(max)(k(u))*ln(3)`
This is shown on the Figure bellow

Distribution of Total Stopping Time

We try to represent the distribution of potential values for Total Stopping Time for a particular number `n`:
`pi_n(k)=Prob(sigma_(oo)(n)=k)`

Parameters

We define an estimated mean:
`overline(sigma_(oo))(n)=sum_(k=0)^(oo)(k*pi_n(k))`
and an estimated variance:
`v(n)=sum_(k=0)^(oo)((k-overline(sigma_(oo))(n))^2*pi_n(k))`

In order to check that model, we need to compare it with real data.
But calculate means and variances is uneasy because we have only one data point for each `n`, and using several points around it would induce a bias.

So we use cumulated mean:
`overline(Sigma_(oo))(n)=(sum_(j=1)^n(sigma_(oo)(j)))/n`
and cumulated variance:
`V(n)=(sum_(j=1)^n(sigma_(oo)(j)-overline(Sigma_(oo))(j))^2)/(n-1)`

The estimated cumulated mean satisfies:
`overline(Sigma_(oo))(n)-n*(del overline(Sigma_(oo))(n))/(del n)=overline(sigma_(oo))(n)`
or, using `u=ln(n)` as parameter:
`overline(Sigma_(oo))(u)-(del overline(Sigma_(oo))(u))/(del u)=overline(sigma_(oo))(u)`

And the estimated cumulated variance satisfies:
`V(n)+n*(del V(n))/(del n)-n^2*((del overline(Sigma_(oo))(n))/(del n))^2=v(n)`
or, using `u=ln(n)` as parameter:
`V(u)+(del V(u))/(del u)-((del overline(Sigma_(oo))(u))/(del u))^2=v(u)`

Model

Taking into account the model we built, for a point with a number of odd steps `o`, a total stopping time `k` and a height `u`, we have:
`o<=o_max(k)=max_o(((k),(o))*1/(3^o)>=1)`

The distribution of `o` for a given `k` is the binomial distribution:
`p_k(o)=r_(k,o)/r_k=(((k),(o))*1/(3^o))/(4/3)^k`

The distribution of `o` for a given `u` is a gamma-like distribution:
`omega(o)=(u-o*ln(3))/ln(2)`
`p_u(o)=(((omega(o)),(o))*1/(3^o))/(sum_(i=0)^(o_max(omega(o)))(((u-i*ln3),(i))*1/(3^i)))`

These distributions are shown on Figure bellow

Mean

Following the route of `n`, in terms of height, we can consider that:

Each Odd Step increases the height by about `ln(3)−ln(2)`:
`n_(i+1)=(3*n_i+1)/2`
`u_(i+1)~~u_i+ln(3)-ln(2)`

Each Even Step decreases the height by `ln(2)`:
`n_(i+1)=n_i/2`
`u_(i+1)=u_i-ln(2)`

And `n_(i+1)` has the same probability `1//2` than `n_i` to belong to each congruence class modulo `2`:

So each step induces an average change in height:
`{:(delta_u,=1/2*(ln(3)-ln(2))+1/2*(-ln(2))),(,=(ln(3)-ln(4))/2):}`

We can write:
`u_k-u_0~~k.(ln(3)-ln(4))/2`

For a complete average route `obrace(n)={n_0=n,cdots,n_(sigma_oo(n))=1}`, `u_(sigma_oo(n)=0`
`\stackrel{~}{u}=sigma_oo(n)*(ln(3)-ln(4))/2`

We have the following result:
`overline(sigma_(oo))(n)=C_1*u`
with
`C_1=2/(2*ln(2)-ln(3))`

Solving the differential equation, it comes:
`overline(Sigma_(oo))(n)=C_1*u-C_1`

Lets us plot the means, calculated versus estimated up to `n=2^28`

Variance

Making numerical identification of the Gamma-like distribution of `o`, and for sufficiently large `u`, we can say that the shape `s` is a linear function of `u`: `s=xi*u`, and the scale `B` is constant. In that case, the mean is `s*B` and the variance is `s*B^2`.

In another hand, `sigma_oo(n)`, `u` and `o` are linked by:
`sigma_oo(n)=(u+o*ln(3))/ln(2)`

it comes:
`{:(overline(sigma_(oo))(n),=(u+s.B*ln(3))/ln(2)),(,=u*((1+xi*B*ln(3))/ln(2))):}`
`{:(v(n),=(s.B^2*ln(3^2))/ln(2)^2),(,=u*((1+xi*B^2*ln(3)^2)/ln(2)^2)),(,=C_2*u):}`

Détails of `C_2` determination are given in the Descriptive Statistics page
`C_2=(C_1^3*ln(3))/4`
`V(n)=C_2*u-C_2+C_1^2`

Lets us plot the cumulated variances, calculated versus estimated up to `n=2^28`