Egad

 

x and z are two separate finite axis marked  of 1/x  and 1/y size.  Let us say they have a  point of symmetric, and can be ordered left and right steps from the the axis. We let z be split along the symmetry with a complement.

Consider all paths through the two axis. The factorial, xz, are the total number of paths, through one partition of z.  Ignor the partition of x, assume it is centered.

If we admit only binomial functions, then the two axis have relative skew, it is the we weighting needed to make an unfair coin fair. Re(z) will be a fraction where the point of symmetry is heads or tails. The total skew is z* skew. When the two axis have balanced two binomials, then the number of coin tosses per binomial is limited by the skew.  That is each binomial get ix and iz coin tosses in proportion needed to keep them both optimally balanced.  We call a skew of one balanced.

Talk about paths from a position on one axis and the other.  The condition must be never.  When Re(z) = z, then the condition is that all paths have two consecutive samples for each new path, in order.   There is only one solution to the Markov 2-tuple.  Everything is multple of two.

I guess we are doing quatenaries, they call it.  Make three or more axis, and try to balance binomials.  x and z must be relative primes because coin tosses are ordered long the axis. If not, two binomials could be merged.

Much easier method then carrying around the imaginaries.   Each time we jump in dimension we have to mark a new point of symmetry in the new axis.  This approach meets all the conditions, should form the masis most most if it.