Larger rates

2, 5, 29

This three color make a big beach ball. The factorial 290 unmerged paths through the error tree. Put all those colors in your color operator. It has the small color appearing twice in a sequence.  

This is a beach ball, consider arranging those colors in concentric circles, three actually. The low rate color firs, red. Then yellow  and blue.   If you took one of the five yellows out, and threw it to the outside ring with blue, then the outside ring has 20 update counts, the middle four and the center two. Then take that total rind and split it into six conics.  That is five exit points per conic.  Then just count path, like a compiler, and merge them.  One be still gets the odd count, green gets a lag, In  cornic will you see two red pending. But that two pending for red only happens in half the conics. The green will see two pending only in one conic, bvut likely when red is odd, where the green ball maintains curvature. So, blue, on the excterior, will have five, except the conic it shares with gree, and then has four.  So blue always has two ending, twice and only red will be pending.

The blue needs to has to have three pending in two coniics.  One it covers when green substiktutes, and the other is a continual deficit, a curvature.

In other words, with five counts per conic, blue is OK with red having two. The condition in Markov is met, two pending, three times  in  each round of  the sequence.  The blue still comes up short, it accumulate a count deficit and surplus around the conic.