From Poisson to Markov

My plot is to substitute Pr(X=k) exactly inot the Markov equation.  Then collect the k! for all three, as well as the lambdas.  Then I find that the terms are met as long as relationship with Log lamdas and k are met to within an integer, I quantize it and prove there is only one set of integer solutions. The constraint is reduced to finding the Markov 'counters' that can replace the Poisson.

One of those, 'Someday', while I hope a young mathematician beats me to it. I am lazy.

First we put the finite N back in Markov

xyz must be the probability that all counters are at one.  For the case of the triple 1,2,5; gthat probability is 1/2 & 1/3 & 1/5, or one out of the n factorial paths. He mupltiplied by N! ^2 and removed N from both sides.

So we reinsert N, and let it go to infinity to get Poisson. We want to see what happens when N tends toward the finite.

So, for each of the factorial error values, 2/3 of the paths were thinned.The counter bump after each valid error update.

So my plot.

Let a,b,c be the Poisson rates and I want the probabilities, or counts, in which k=1, the number error upsaets pending is one, for all three together versus each twice at a time. I get a very simply equation, my only transcendental is an exponent is, Euler. But I am flooring the equation to rational fractions, and need only know the natural exponent to the nearest one/half. I end up with a second degree equation in a,b,c which should be Markov again.   How does that help? Not sure yet, just a proof a different algorithm exists.

But this does let me run a flat earth version of Markov.  One neat thing I can do is set the highest counter to some value, then simplify the condition that k=1. At that point the lowest rate will be either zero, od some positive number, thus proving there is a Markov triplre for any integer c in:(a,b,c) Doing so might get me a prize, but it is still work.