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

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 zeros, the first rows of the square tables for form the following triangle of numbers, which is is symmetric up to except for two rows. The first terms of the rows of this triangle appear to be the number of binary Lyndon words of length A001037 shifted by three and the last terms of the rows appear to be the absolute values of the sequence A038063 shifted by two.