Eulerian Numbers versus Stirling Numbers of the First Kind

In a permutation , an ascent is a pair with . The Eulerian number counts the number of permutations of with exactly ascents, . Alternatively, counts the number of permutations of with exactly ' permutation runs, . For example, the permutation has the three ascents , , and and the two runs and .

The unsigned Stirling number of the first kind counts the number of permutations of whose cycle decomposition has cycles. For example, the permutation is the mapping , , , , , so its cycle decomposition is with four cycles.

Both types of numbers count the permutations of in different ways. Define the number to count the number of permutations that have cycles and runs. This Demonstration lays out these numbers in a square; the row sums are and the column sums are ; the sums of either of those is .

The 1 in the first column comes from the identity permutation that has cycles row) and one run (first column).

The 1 in the last column comes from the permutation . It has runs of length 1. Its cycle decomposition is for even and for odd, so has cycles.

If a permutation has many cycles, they chain many elements together into longer but fewer runs, which explains the zeros in the lower-right part of the tables.

Let us see why the two nonzero entries in the second-to-last row for are and . A permutation with cycles consists of one transposition and singletons (fixed points); there are such permutations. If the pair in the transposition are neighbors, there are two runs; there are ways to form a neighboring pair. If the pair are not neighbors, there are three runs and such pairs.

Ignoring the first two squares and the zeros at the beginning and end, the first rows form the following triangle of numbers, which is is symmetric up to except for two rows. The rows appear to start with the number of binary Lyndon words of length (A001037) and appear to end with a sequence directly related to A038063.