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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
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 1,3
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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.
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LINKS
| Alois P. Heinz, Rows n = 1..100, flattened
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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
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EXAMPLE
| 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
| with (combinat):
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
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CROSSREFS
| Cf. A039810, A039814, A126350, A054654, A126353.
Sequence in context: A056242 A128718 A112358 * A157011 A092821 A110552
Adjacent sequences: A126348 A126349 A126350 * A126352 A126353 A126354
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KEYWORD
| nonn,tabl
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AUTHOR
| Thomas Wieder (thomas.wieder(AT)t-online.de), Dec 29 2006
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