OFFSET
0,8
COMMENTS
Let a=a(1)a(2)...a(2*n) be a permutation of [2*n] written in one line notation i.e. written as a word. T(n,k) is the number of ways to factor the permutations of [2*n] 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.
FORMULA
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.
From Seiichi Manyama, Apr 14 2026: (Start)
T(n,k) = (2*n)! * [x^(2*n)] sec(x)^k.
T(0,k) = 1 and T(n,k) = k*(k+1) * T(n-1,k+2) - k^2 * T(n-1,k) for n > 0. (End)
EXAMPLE
Square array begins:
1, 1, 1, 1, 1, 1, ...
0, 1, 2, 3, 4, 5, ...
0, 5, 16, 33, 56, 85, ...
0, 61, 272, 723, 1504, 2705, ...
0, 1385, 7936, 25953, 64256, 134185, ...
0, 50521, 353792, 1376643, 3963904, 9451805, ...
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.
MATHEMATICA
nn = 5; B[n_] := (2 n)!/2^n; e[x_] := Sum[x^n/B[n], {n, 0, nn}];
Table[Table[B[n], {n, 0, nn}] PadRight[CoefficientList[Series[e[-x]^-k, {x, 0, nn}], x], nn + 1], {k, 0, nn}] // Transpose // Grid
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Geoffrey Critzer, Feb 07 2021
STATUS
approved
