%I #8 Feb 07 2021 21:09:57

%S 1,1,0,1,1,0,1,2,5,0,1,3,16,61,0,1,4,33,272,1385,0,1,5,56,723,7936,

%T 50521,0,1,6,85,1504,25953,353792,2702765,0,1,7,120,2705,64256,

%U 1376643,22368256,199360981,0

%N Square array read by descending antidiagonals. T(n,k) is the number of ways to factor a permutation of [2n] into exactly k good factors, n>=0, k>=0.

%C Let a=a(1)a(2)...a(2n) be a permutation of [2n] written in one line notation i.e. written as a word. T(n,k) is the number of ways to factor the permutations of [2n] into k factors [a(1)...a(j_1)] [a(j_1+1)... a(j_2)]... [a(j_{k-1}+1) ... a(j_k=2n)] so that each factor has even size and descent set {2,4,6,...,j_i - 2} for i ={1,2,...,k}. In other words, the factors are the up-down permutations enumerated in A000364. The factors may be empty.

%F T(n,k) = Z(P_n,-k) where Z(P_n,k) is the zeta polynomial of an n-interval in the binomial poset of even sized subsets of the positive integers.

%e 1, 1, 1, 1, 1, 1, ...

%e 0, 1, 2, 3, 4, 5, ...

%e 0, 5, 16, 33, 56, 85, ...

%e 0, 61, 272, 723, 1504, 2705, ...

%e 0, 1385, 7936, 25953, 64256, 134185, ...

%e 0, 50521, 353792, 1376643, 3963904, 9451805, ...

%e T(2,2) = 16. There is 1 factorization into 2 good factors of the permutation 1234=[12][34]. Each of the permutations 1324,1423,2314,2413,3412 can be factored in 3 good ways. For example 1324=[][1324]=[13][24]=[1324][]. So T(2,2) = 1 + 3(5) = 16.

%t nn = 5; B[n_] := (2 n)!/2^n; e[x_] := Sum[x^n/B[n], {n, 0, nn}];

%t Table[Table[B[n], {n, 0, nn}] PadRight[CoefficientList[Series[e[-x]^-k, {x, 0, nn}], x], nn + 1], {k, 0, nn}] // Transpose // Grid

%Y Cf. A000364 (column k=1), A000182 (column k=2).

%K nonn,tabl

%O 0,8

%A _Geoffrey Critzer_, Feb 07 2021