A320618 -id:A320618 - cp's OEIS frontend

A276891 Number T(n,k) of ordered set partitions of [n] where k is minimal such that for each block b the smallest integer interval containing b has at most k elements; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

1, 0, 1, 0, 2, 1, 0, 6, 4, 3, 0, 24, 20, 18, 13, 0, 120, 114, 118, 114, 75, 0, 720, 750, 878, 924, 870, 541, 0, 5040, 5616, 7224, 8152, 8760, 7818, 4683, 0, 40320, 47304, 65514, 79682, 90084, 94560, 81078, 47293, 0, 362880, 443400, 652446, 845874, 998560, 1135776, 1148016, 954474, 545835

Offset: 0

Alois P. Heinz, Sep 21 2016

			Triangle T(n,k) begins:
  1;
  0,     1;
  0,     2,     1;
  0,     6,     4,     3;
  0,    24,    20,    18,    13;
  0,   120,   114,   118,   114,    75;
  0,   720,   750,   878,   924,   870,   541;
  0,  5040,  5616,  7224,  8152,  8760,  7818,  4683;
  0, 40320, 47304, 65514, 79682, 90084, 94560, 81078, 47293;
  ...

Alois P. Heinz, Rows n = 0..21, flattened

Columns k=0-10 give: A000007, A000142 (for n>0), A320615, A320616, A320617, A320618, A320619, A320620, A320621, A320622, A320623.
Row sums give: A000670.
Main diagonal gives A000670(n-1) for n>0.
T(2n,n) gives A276892.
Cf. A276727, A276890.

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)]))):
T:= (n, k)-> A(n, k) -`if`(k=0, 0, A(n, k-1)):
seq(seq(T(n, k), k=0..n), n=0..10);

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]]]]; T [n_, k_] := A[n, k] - If[k == 0, 0, A[n, k - 1]]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 04 2017, translated from Maple *)

T(n,k) = A276890(n,k) - A276890(n,k-1) for k>0, T(n,0) = A000007(n).

A320555 Number of set partitions of [n] such that for each block b the smallest integer interval containing b has at most five elements and for at least one block c the smallest integer interval containing c has exactly five elements.

Original entry on oeis.org

15, 64, 201, 585, 1741, 5375, 16355, 48601, 141921, 410425, 1182828, 3398411, 9728692, 27745449, 78861484, 223573925, 632578393, 1786856056, 5039984789, 14197033194, 39945491361, 112282665839, 315352029653, 885048266680, 2482371076351, 6958712870273

Offset: 5

Alois P. Heinz, Oct 15 2018

Alois P. Heinz, Table of n, a(n) for n = 5..1000

Column k=5 of A276727.

Cf. A276720, A276721, A320618.

b:= proc(n, m, l) option remember; `if`(n=0, 1,
      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`(n=0, 1, `if`(k<2, k, b(n, 0, [0$(k-1)]))):
a:= n-> (k-> A(n, k) -`if`(k=0, 0, A(n, k-1)))(5):
seq(a(n), n=5..50);

b[n_, m_, l_List] := b[n, m, l] = If[n == 0, 1, 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[n == 0, 1, If[k < 2, k, b[n, 0, Array[0 &, k - 1]]]];
a[n_] := With[{k = 5}, A[n, k] - If[k == 0, 0, A[n, k - 1]]];
a /@ Range[5, 35] (* Jean-François Alcover, Dec 08 2020, after Alois P. Heinz *)

a(n) = A276721(n) - A276720(n).

cp's OEIS Frontend

A276891 Number T(n,k) of ordered set partitions of [n] where k is minimal such that for each block b the smallest integer interval containing b has at most k elements; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

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