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A145324
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Triangle read by rows: coefficients of 1; 1(X+2); 1(X+2)(X+3); 1(X+2)(X+3)(X+4); ....
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11
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1, 1, 2, 1, 5, 6, 1, 9, 26, 24, 1, 14, 71, 154, 120, 1, 20, 155, 580, 1044, 720, 1, 27, 295, 1665, 5104, 8028, 5040, 1, 35, 511, 4025, 18424, 48860, 69264, 40320, 1, 44, 826, 8624, 54649, 214676, 509004, 663696, 362880, 1, 54, 1266, 16884, 140889, 761166
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OFFSET
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1,3
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COMMENTS
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The last number of row n is n!.
Essentially the triangle given by [1, 0, 1, 0, 1, 0, 1, 0, 1, ...] DELTA [2, 1, 3, 2, 4, 3, 5, 4, 6, 5, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 09 2008
T(n+1,k+1) = a_k(2,3,...,n+1), n >= 0, k = 0..n, with the elementary symmetric function a_k(x[1],x[2],...,x[n]), with a_0(0):=1. E.g., a_2(2,3,4) = 2*3 + 2*4 + 3*4 = 26 = T(4,3). - Wolfdieter Lang, Oct 24 2011
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LINKS
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FORMULA
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T(n,k) = Sum_{m=0..k-1} (-1)^m*|s(n+1, n+2-k+m)|, n >= 1, k = 1..n, with the Stirling numbers of the first kind s(n,k) = A048994(n,k). - Wolfdieter Lang, Oct 24 2011
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EXAMPLE
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n\k 1 2 3 4 5 6 7 ...
1: 1
2: 1 2
3: 1 5 6
4: 1 9 26 24
5: 1 14 71 154 120
6: 1 20 155 580 1044 720
7: 1 27 295 1665 5104 8028 5040
...
T(4,3)= 26 = |s(5,3)| - |s(5,4)| + |s(5,5)| = 35 - 10 + 1.
(End)
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MAPLE
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A145324 := proc(n, k) coeftayl( 1*mul(x+i, i=2..n), x=0, n-k) ; end: for n from 1 to 11 do for k from 1 to n do printf("%d, ", A145324(n, k)) ; od: od: # R. J. Mathar, Oct 10 2008
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MATHEMATICA
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Table[Reverse[CoefficientList[Product[x+j, {j, 2, k}], x]], {k, 1, 15}] // Flatten (* Robert A. Russell, Sep 29 2018 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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