# Is there a planar network whose path give a TNN matrix whose entries are Eulerian numbers?

It is well-known that for each planar graph, the number of paths from each source to each sink will give rise to a totally non-negative matrix. However, did anyone ever come up with a planar network whose paths give a matrix whose entries are Eulerian numbers?

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Do you need a nondegenerate example? There is one with all the sinks coinciding. – Gjergji Zaimi Apr 19 '12 at 10:23
I was hoping to get a nondegenerate example. But sure, I would love to see even a degenerated one thanks!!! – Percy Apr 19 '12 at 14:19

There is a Bratelli diagram which satisfies this. Consider $\mathbb N^2$ as a graph where the edges are given by k parallel directed edges from $(n,k)$ to $(n+1,k)$, and $n-k+1$ parallel directed edges from $(n,k)$ to $(n+1,k+1)$. Here the number of paths from $(1,1)$ (the unique source) to any vertex $(n,k)$ is the Eulerian number $A(n,k)$, the number of permutations in $S_n$ with exactly $k$ ascents. For a picture see fig.1 in "Random permutations and unique fully supported ergodicity for the Euler adic transformation" by Frick and Petersen. This can be proven quickly using the recurrence $$A(n, k) = (k+1) A(n-1, k) + (n-k+1) A(n-1, k-1).$$
Somewhat similarly one can give a directed graph where the number of paths from the source to any vertex in the graph is equal to an Euler number $E(n,k)$, which counts the number of alternating permutations of $S_n$ which start with $k$. For a picture of how this looks like look at fig.5 in "Generating alternating permutations lexicographically" by Bauslaugh and Ruskey.