Markov condition

For some dimension equal relative primes. The deviation counts are finite and the relative primes need only count through them twice, in total. Boltzmann is (n-1)/n;  I think, that is a lot of path reduction, or undersampling.

On any Markov N-Tuple tree, going up the tree means having another finite, but larger, Bayes set N of deviations total.  There is a limit, says Hurwitz.  As a limit there is residual round off error, the Markov gap?

The Generalized Nyquist-Shannon tells us the sample space bandwidth per dimension. Everything else follows, except kinetic energy, modeled as the unfinished coin toss, needs some work.

The deviation arrivals are relatively prime, and the total number of combinations is x^x y^y.....

This divides up a base two axis, and the last number much larger, several digits down in a twos system.   But, about their two digit, they should be binomial distributed.

Then the total probability of counting twice is four, if we count all [paths four times..  But they cannot over lap on counting twice.   As the dimension expands so does the gap, the differential exponent separating them as binomial distributions.   The exponent is set by the number of coin tosses, cannot toss two coins at once.