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? 



Is there a planar network whose path give a TNN matrix whose entries are Eulerian numbers?
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](https://en.wikipedia.org/wiki/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"](https://projecteuclid.org/journals/annales-de-linstitut-henri-poincare-probabilites-et-statistiques/volume-44/issue-5/Random-permutations-and-unique-fully-supported-ergodicity-for-the-Euler/10.1214/07-AIHP133.full) 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](https://en.wikipedia.org/wiki/Alternating_permutation) of $S\_n$ which start with $k$. For a picture of how this looks like look at fig.5 in ["Generating alternating permutations lexicographically"](http://www.springerlink.com/content/xx43404n2t4388p0/) by Bauslaugh and Ruskey.


