A276890 Number A(n,k) of ordered set partitions of [n] such that for each block b the smallest integer interval containing b has at most k elements; square array A(n,k), n>=0, k>=0, read by antidiagonals.
1, 1, 0, 1, 1, 0, 1, 1, 2, 0, 1, 1, 3, 6, 0, 1, 1, 3, 10, 24, 0, 1, 1, 3, 13, 44, 120, 0, 1, 1, 3, 13, 62, 234, 720, 0, 1, 1, 3, 13, 75, 352, 1470, 5040, 0, 1, 1, 3, 13, 75, 466, 2348, 10656, 40320, 0, 1, 1, 3, 13, 75, 541, 3272, 17880, 87624, 362880, 0
Offset: 0
Examples
Square array A(n,k) begins: 1, 1, 1, 1, 1, 1, 1, 1, ... 0, 1, 1, 1, 1, 1, 1, 1, ... 0, 2, 3, 3, 3, 3, 3, 3, ... 0, 6, 10, 13, 13, 13, 13, 13, ... 0, 24, 44, 62, 75, 75, 75, 75, ... 0, 120, 234, 352, 466, 541, 541, 541, ... 0, 720, 1470, 2348, 3272, 4142, 4683, 4683, ... 0, 5040, 10656, 17880, 26032, 34792, 42610, 47293, ...
Links
- Alois P. Heinz, Antidiagonals n = 0..42, flattened
Crossrefs
Programs
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Maple
b:= proc(n, m, l) option remember; `if`(n=0, m!, add(b(n-1, max(m, j), [subsop(1=NULL, l)[], `if`(j<=m, 0, j)]), j={l[], m+1} minus {0})) end: A:= (n, k)-> `if`(k=0, `if`(n=0, 1, 0), `if`(k=1, n!, b(n, 0, [0$(k-1)]))): seq(seq(A(n, d-n), n=0..d), d=0..12);
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Mathematica
b[n_, m_, l_List] := b[n, m, l] = If[n == 0, m!, Sum[b[n - 1, Max[m, j], Append[ReplacePart[l, 1 -> Nothing], If[j <= m, 0, j]]], {j, Append[l, m + 1] ~Complement~ {0}}]]; A[n_, k_] := If[k==0, If[n==0, 1, 0], If[k==1, n!, b[n, 0, Array[0&, k-1]]]]; Table[A[n, d-n], {d, 0, 12}, {n, 0, d}] // Flatten (* Jean-François Alcover, Jan 06 2017, after Alois P. Heinz *)
Formula
A(n,k) = Sum_{j=0..k} A276891(n,j).
Comments