A126351 Triangle read by rows: matrix product of the Stirling numbers of the second kind with the binomial coefficients.

1, 1, 2, 1, 5, 4, 1, 9, 19, 8, 1, 14, 55, 65, 16, 1, 20, 125, 285, 211, 32, 1, 27, 245, 910, 1351, 665, 64, 1, 35, 434, 2380, 5901, 6069, 2059, 128, 1, 44, 714, 5418, 20181, 35574, 26335, 6305, 256, 1, 54, 1110, 11130, 58107, 156660, 204205, 111645, 19171, 512

Offset: 1

Author

Thomas Wieder, Dec 29 2006

Keywords

• nonn
• tabl

Comments

Many well-known integer sequences arise from such a matrix product of combinatorial coefficients. In the present case we have as the first row A000079 = the powers of two = 2^n. As the second row we have A001047 = 3^n - 2^n. As the column sums we have 1,3,10,37,151,674,3263,17007,94828 we have A005493 = number of partitions of [n+1] with a distinguished block.

Examples

			Matrix begins:
1, 2, 4,  8, 16,  32,   64,  128,   256, ... A000079
0, 1, 5, 19, 65, 211,  665, 2059,  6305, ... A001047
0, 0, 1,  9, 55, 285, 1351, 6069, 26335, ... A016269
0, 0, 0,  1, 14, 125,  910, 5901, 35574, ... A025211
0, 0, 0,  0,  1,  20,  245, 2380, 20181, ...
0, 0, 0,  0,  0,   1,   27,  434,  5418, ...
0, 0, 0,  0,  0,   0,    1,   35,   714, ...
0, 0, 0,  0,  0,   0,    0,    1,    44, ...
0, 0, 0,  0,  0,   0,    0,    0,     1, ...
Triangle begins:
1;
1,  2;
1,  5,  4;
1,  9, 19,  8;
1, 14, 55, 65, 16;
		

Links

• Alois P. Heinz, Rows n = 1..100, flattened

Crossrefs

Cf. A039810, A039814, A126350, A054654, A126353.

Programs

Maple

T:= (n, k)-> add(binomial(n-1, i-1) *Stirling2(i, n+1-k), i=1..n):
seq(seq(T(n, k), k=1..n), n=1..10);  # Alois P. Heinz, Sep 29 2011

Mathematica

T[n_, k_] := Sum[Binomial[n-1, i-1]*StirlingS2[i, n+1-k], {i, 1, n}]; Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jan 08 2016, after Alois P. Heinz *)

Formula

(In Maple notation:) Matrix product B.A of matrix A[i,j]:=binomial(j-1,i-1) with i = 1 to p+1, j = 1 to p+1, p=8 and of matrix B[i,j]:=stirling2(j,i) with i from 1 to d, j from 1 to d, d=9.

T(n,k) = Sum_{i=1..n} C(n-1,i-1) * Stirling2(i, n+1-k). - Alois P. Heinz, Sep 29 2011