A355260 Triangle read by rows, T(n, k) = Bell(k) * |Stirling1(n, k)|.

1, 0, 1, 0, 1, 2, 0, 2, 6, 5, 0, 6, 22, 30, 15, 0, 24, 100, 175, 150, 52, 0, 120, 548, 1125, 1275, 780, 203, 0, 720, 3528, 8120, 11025, 9100, 4263, 877, 0, 5040, 26136, 65660, 101535, 101920, 65366, 24556, 4140, 0, 40320, 219168, 590620, 1009260, 1167348, 920808, 478842, 149040, 21147

Offset

0,6

Links

Table of n, a(n) for n=0..54.

Formula

T(n, k) = n! * [y^k] [x^n] exp(1/(1 - x)^y - 1).

T(n, k) = Bell(k)*Bell_{n, k}(A000142), where Bell_{n, k}(S) are the partial Bell polynomials mapped on the sequence S; here S are the factorial numbers. See the Mathematica program.

T(n, k) = A000110(k) * A132393(n, k).

Example

Triangle T(n, k) begins:
[0] 1;
[1] 0,   1;
[2] 0,   1,    2;
[3] 0,   2,    6,    5;
[4] 0,   6,   22,   30,   15;
[5] 0,  24,  100,  175,  150,    52;
[6] 0, 120,  548, 1125, 1275,   780,  203;
[7] 0, 720, 3528, 8120, 11025, 9100, 4263, 877;

Maple

Bell := n -> combinat[bell](n):
T := (n, k) -> Bell(k)*abs(Stirling1(n, k)):
seq(seq(T(n, k), k = 0..n), n = 0..9);
# Alternative:
egf := exp(1/(1 - x)^y - 1): ser := series(egf, x, 32):
cfx := n -> coeff(ser, x, n):
seq(seq(n!*coeff(cfx(n), y, k), k = 0..n), n = 0..8);

Mathematica

(* Utility function, extracts the lower triangular part of a square matrix. *)
TriangularForm[T_] := Table[Table[T[[n, k]], {k, 1, n}], {n, 1, Dimensions[T][[1]]}];
(* The actual calculation: *)
r := 9; R := Range[0, r];
T := Table[BellB[k] BellY[n, k, R!], {n, R}, {k, R}];
T // TriangularForm // Flatten

Crossrefs

Cf. A000262 (row sums), A033999 (alternating row sums), A000110 (main diagonal), A000142 (column 1).

Cf. A132393, A355267.

Sequence in context: A327116 A157491 A094385 * A291799 A295027 A225479

Adjacent sequences: A355257 A355258 A355259 * A355261 A355262 A355263

Keyword

nonn,tabl

Author

Peter Luschny, Jul 06 2022

Status

approved