A287213 Number T(n,k) of set partitions of [n] such that the maximal absolute difference between consecutive elements within a block equals k; triangle T(n,k), n>=0, 0<=k<=max(n-1,0), read by rows.

1, 1, 1, 1, 1, 3, 1, 1, 7, 5, 2, 1, 15, 18, 13, 5, 1, 31, 57, 61, 38, 15, 1, 63, 169, 248, 215, 129, 52, 1, 127, 482, 935, 1061, 836, 495, 203, 1, 255, 1341, 3368, 4835, 4789, 3573, 2108, 877, 1, 511, 3669, 11777, 20973, 25430, 22986, 16657, 9831, 4140

Offset: 0

Author

Alois P. Heinz, May 21 2017

Keywords

• nonn
• tabf

Comments

The maximal absolute difference is assumed to be zero if there is no block with consecutive elements.

T(n,k) is defined for all n,k >= 0. The triangle contains only the positive terms. T(n,k) = 0 if k>=n and k>0.

Examples

			T(4,0) = 1: 1|2|3|4.
T(4,1) = 7: 1234, 123|4, 12|34, 12|3|4, 1|234, 1|23|4, 1|2|34.
T(4,2) = 5: 124|3, 134|2, 13|24, 13|2|4, 1|24|3.
T(4,3) = 2: 14|23, 14|2|3.
Triangle T(n,k) begins:
  1;
  1;
  1,   1;
  1,   3,   1;
  1,   7,   5,   2;
  1,  15,  18,  13,    5;
  1,  31,  57,  61,   38,  15;
  1,  63, 169, 248,  215, 129,  52;
  1, 127, 482, 935, 1061, 836, 495, 203;
		

Links

• Alois P. Heinz, Rows n = 0..23, flattened
• Wikipedia, Partition of a set

Crossrefs

Columns k=0-10 give: A000012, A000225(n-1), A258109, A294052, A294053, A294054, A294055, A294056, A294057, A294058, A294059.

Row sums and T(n+2,n+1) give A000110.

T(2n,n) gives A294024.

Cf. A287214, A287215, A287416, A287640.

Programs

Maple

b:= proc(n, k, l) option remember; `if`(n=0, 1, b(n-1, k, map(x->
      `if`(x-n>=k, [][], x), [l[], n]))+add(b(n-1, k, sort(map(x->
      `if`(x-n>=k, [][], x), subsop(j=n, l)))), j=1..nops(l)))
    end:
A:= (n, k)-> b(n, min(k, n-1), []):
T:= (n, k)-> A(n, k)-`if`(k=0, 0, A(n, k-1)):
seq(seq(T(n, k), k=0..max(n-1, 0)), n=0..12);

Mathematica

b[0, , ] = 1; b[n_, k_, l_] := b[n, k, l] =b[n - 1, k, If[# - n >= k, Nothing, #]& /@ Append[l, n]] + Sum[b[n - 1, k, Sort[If[# - n >= k, Nothing, #]& /@ ReplacePart[l, j -> n]]], {j, 1, Length[l]}];
A[n_, k_] := b[n, Min[k, n - 1], {}];
T[n_, k_] :=  A[n, k] - If[k == 0, 0, A[n, k - 1]];
Table[Table[T[n, k], {k, 0, Max[n - 1, 0]}], {n, 0, 12}] // Flatten (* Jean-François Alcover, Apr 30 2018, after Alois P. Heinz *)

Formula

T(n,k) = A287214(n,k) - A287214(n,k-1) for k>0, T(n,0) = 1.