Multiple of M in the Markov M-tuple

 Markov M-tuples are about conserving histograms squares when combining finite histograms.

In the 3-tuple, x,y,z are the number of spots in three different histograms.

x^2 is the number oif histogram spots needed for any event in x to happen twice. xyz is the number of histogram spots needed if each histogrram happen once.

The three, x,y,z must be relative primes.  Otherwise xyz ends up being a power of the common factor, by recursion.  Consider a two dimensional histogram of total size yz.  Is their a one dimensional histogram which contains the same information? Yes, if I can find a two dimensional ordering, I need two seep and discrete angle across the two dimensional surface which some finite delta radius.   This is the same as graphing two geodesic ratios around Lie Graph paper.

These are relative prime coin tosses.  They are ordered. Consider:

(1,2..y)*(1,2,..z).  There is no common term, this distribution has yz histogram points. E#ach, by them selves, represent a positive definite cumulative distribution.  So, each spot, individually, as a norm, i/y or i/y, i an increasing integer set.  Thus, the only multiple is yz. Each of the ij must be unique and nor increases with i and j thus increasing with ij:  As long as the finite ratio x/y always tends to maximize entropy.

Consider one individual histogram spot. say one on the x axis.  The path length through the factorial xyz is three. No short cut symmetries because of relative primeness. If any of x is counted are the start of a path, then the next path cannot start with x, else x is a multiple of three. The paths, in total are partition into distinct bundles.   

Here is another clue.  When x,y,z maximally approximate a binomial, individually, then they are maximally independent, and within an integer can be approximated as three Gaussian arrivals and meet thr Markov condition.  Then, if we include ordering, the total number of integer spots available in tossing the coins twice for each of x,y,z will be a multiple of three, the dimensionality.  This means the three dimensions are at optimum, these are optimum relative primes in Markov 3 tuples tuple.

Here is a model:

We have 3 layers in a 3 tuple, 4 in a four tuple and so one.  But our layers are geodesic, fractional approximations to Pi. The fraction selected is the one that minimizes the residual in round off error, in this model.  I am approximating flat earth, in that my mechanics force circular motion, so I get Abvogadro count rather than proton count.  I went through a deal of effort in sanding my circular mold and raising temperatures.

I digress. Consider a path through the insulation layers in geodesic steps. One of my histogram spots goes a lot further down then the others, depending on radius.  The farther down a step takes you, the fewer steps you take, ordering is preserved.  These can be arranged as distributions about the circumference, in steps of three, when optimized.  So, waddya know, Plack's number is really the number of layers. You can count paths by three, the right side of Markov, the number of paths is segregated by starting color.

Note also, for any Markov m-tuple the smallest unique elements are:

1,m-1,y,z,....

So, the 3 tuple condition, 3xyz is the first unique 4 tuple with 3 being the smallest 4 tuple.

Each jump up the m-tuple level makes the previous prime the smallest four tuple..  

What happens when we hit 3^3 in a 3 tuple? I dunno.  But the minimum N needs to make unique M+1 bundle of paths goes higher in the M+1 Tree.  Unique means the M+1 tuple cannot have multiple primes which can be composed into a M tuple in the lower tree.  

So, in finale, there is a relationship between relative primes and the optimum set of histogram spots.  This is a general spectral estimation result, an expansion to any integer dimension.