A124327 Triangle read by rows: T(n,k) is the number of partitions of the set {1,2,...,n} such that the sum of the least entries of the blocks is k (1<=k<=n*(n+1)/2).

1, 1, 0, 1, 1, 0, 2, 1, 0, 1, 1, 0, 4, 2, 1, 3, 2, 1, 0, 1, 1, 0, 8, 4, 2, 10, 6, 7, 2, 5, 3, 2, 1, 0, 1, 1, 0, 16, 8, 4, 29, 19, 21, 14, 23, 14, 18, 10, 7, 7, 5, 3, 2, 1, 0, 1, 1, 0, 32, 16, 8, 85, 56, 64, 42, 101, 62, 75, 69, 47, 54, 38, 38, 24, 23, 10, 13, 7, 5, 3, 2, 1, 0, 1, 1, 0, 64, 32, 16

Offset: 1

Author

Emeric Deutsch, Oct 31 2006

Keywords

• nonn
• tabf

Comments

Row n has n(n+1)/2 terms. Row sums yield the Bell numbers (A000110). T(n,1)=1; T(n,2)=0; T(n,3)=2^(n-2). for n>=2; T(n,4)=2^(n-3) for n>=3; T(n,5)=2^(n-4) for n>=4.

Examples

			T(4,7) = 2 because we have 13|2|4 and 1|23|4.
Triangle starts:
  1;
  1, 0,  1;
  1, 0,  2, 1, 0,  1;
  1, 0,  4, 2, 1,  3,  2,  1,  0,  1;
  1, 0,  8, 4, 2, 10,  6,  7,  2,  5,  3,  2,  1, 0, 1;
  1, 0, 16, 8, 4, 29, 19, 21, 14, 23, 14, 18, 10, 7, 7, 5, 3, 2, 1, 0, 1;
  ...
		

Links

• Alois P. Heinz, Rows n = 1..40, flattened

Crossrefs

Cf. A000110, A000217, A124325, A367955.

Antidiagonal sums give A365821.

Row maxima give A365903.

T(n,n) gives A368204.

Programs

Maple

Q[1]:=t*s: for n from 2 to 8 do Q[n]:=expand(s*diff(Q[n-1],s)+t^n*s*Q[n-1]) od: for n from 1 to 8 do P[n]:=sort(subs(s=1,Q[n])) od: for n from 1 to 8 do seq(coeff(P[n],t,k),k=1..n*(n+1)/2) od; # yields sequence in triangular form

Mathematica

Q[1, t_, s_] := t s;
Q[n_, t_, s_] := Q[n, t, s] = s D[Q[n-1, t, s], s] + s t^n Q[n-1, t, s] // Expand;
P[n_, t_] := Q[n, t, s] /. s -> 1;
T[n_] := Rest@CoefficientList[P[n, t], t];
Table[T[n], {n, 1, 8}] // Flatten (* Jean-François Alcover, Jun 10 2024 *)

Formula

The generating polynomial of row n is P(n,t)=Q(n,t,1), where Q(n,t,s)=s*dQ(n-1,t,s)/ds + st^n*Q(n-1,t,s); Q(1,t,s)=ts.

Sum_{k=1..n*(n+1)/2} k * T(n,k) = A124325(n+1). - Alois P. Heinz, Dec 05 2023