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A126351
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Triangle read by rows: matrix product of the Stirling numbers of the second kind with the binomial coefficients.
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5
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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
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OFFSET
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1,3
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COMMENTS
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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.
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LINKS
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FORMULA
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(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
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EXAMPLE
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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;
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MAPLE
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T:= (n, k)-> add(binomial(n-1, i-1) *Stirling2(i, n+1-k), i=1..n):
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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