%I #22 Jan 08 2016 16:01:02
%S 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,
%T 1351,665,64,1,35,434,2380,5901,6069,2059,128,1,44,714,5418,20181,
%U 35574,26335,6305,256,1,54,1110,11130,58107,156660,204205,111645,19171,512
%N Triangle read by rows: matrix product of the Stirling numbers of the second kind with the binomial coefficients.
%C 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.
%H Alois P. Heinz, <a href="/A126351/b126351.txt">Rows n = 1..100, flattened</a>
%F (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.
%F T(n,k) = Sum_{i=1..n} C(n-1,i-1) * Stirling2(i, n+1-k). - _Alois P. Heinz_, Sep 29 2011
%e Matrix begins:
%e 1, 2, 4, 8, 16, 32, 64, 128, 256, ... A000079
%e 0, 1, 5, 19, 65, 211, 665, 2059, 6305, ... A001047
%e 0, 0, 1, 9, 55, 285, 1351, 6069, 26335, ... A016269
%e 0, 0, 0, 1, 14, 125, 910, 5901, 35574, ... A025211
%e 0, 0, 0, 0, 1, 20, 245, 2380, 20181, ...
%e 0, 0, 0, 0, 0, 1, 27, 434, 5418, ...
%e 0, 0, 0, 0, 0, 0, 1, 35, 714, ...
%e 0, 0, 0, 0, 0, 0, 0, 1, 44, ...
%e 0, 0, 0, 0, 0, 0, 0, 0, 1, ...
%e Triangle begins:
%e 1;
%e 1, 2;
%e 1, 5, 4;
%e 1, 9, 19, 8;
%e 1, 14, 55, 65, 16;
%p T:= (n, k)-> add(binomial(n-1, i-1) *Stirling2(i, n+1-k), i=1..n):
%p seq(seq(T(n, k), k=1..n), n=1..10); # _Alois P. Heinz_, Sep 29 2011
%t 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_ *)
%Y Cf. A039810, A039814, A126350, A054654, A126353.
%K nonn,tabl
%O 1,3
%A _Thomas Wieder_, Dec 29 2006