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A276890
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Number A(n,k) of ordered set partitions of [n] such that for each block b the smallest integer interval containing b has at most k elements; square array A(n,k), n>=0, k>=0, read by antidiagonals.
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12
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1, 1, 0, 1, 1, 0, 1, 1, 2, 0, 1, 1, 3, 6, 0, 1, 1, 3, 10, 24, 0, 1, 1, 3, 13, 44, 120, 0, 1, 1, 3, 13, 62, 234, 720, 0, 1, 1, 3, 13, 75, 352, 1470, 5040, 0, 1, 1, 3, 13, 75, 466, 2348, 10656, 40320, 0, 1, 1, 3, 13, 75, 541, 3272, 17880, 87624, 362880, 0
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OFFSET
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0,9
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COMMENTS
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LINKS
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FORMULA
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A(n,k) = Sum_{j=0..k} A276891(n,j).
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EXAMPLE
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Square array A(n,k) begins:
1, 1, 1, 1, 1, 1, 1, 1, ...
0, 1, 1, 1, 1, 1, 1, 1, ...
0, 2, 3, 3, 3, 3, 3, 3, ...
0, 6, 10, 13, 13, 13, 13, 13, ...
0, 24, 44, 62, 75, 75, 75, 75, ...
0, 120, 234, 352, 466, 541, 541, 541, ...
0, 720, 1470, 2348, 3272, 4142, 4683, 4683, ...
0, 5040, 10656, 17880, 26032, 34792, 42610, 47293, ...
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MAPLE
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b:= proc(n, m, l) option remember; `if`(n=0, m!,
add(b(n-1, max(m, j), [subsop(1=NULL, l)[],
`if`(j<=m, 0, j)]), j={l[], m+1} minus {0}))
end:
A:= (n, k)-> `if`(k=0, `if`(n=0, 1, 0),
`if`(k=1, n!, b(n, 0, [0$(k-1)]))):
seq(seq(A(n, d-n), n=0..d), d=0..12);
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MATHEMATICA
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b[n_, m_, l_List] := b[n, m, l] = If[n == 0, m!, Sum[b[n - 1, Max[m, j], Append[ReplacePart[l, 1 -> Nothing], If[j <= m, 0, j]]], {j, Append[l, m + 1] ~Complement~ {0}}]]; A[n_, k_] := If[k==0, If[n==0, 1, 0], If[k==1, n!, b[n, 0, Array[0&, k-1]]]]; Table[A[n, d-n], {d, 0, 12}, {n, 0, d}] // Flatten (* Jean-François Alcover, Jan 06 2017, after Alois P. Heinz *)
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CROSSREFS
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Columns k=0-10: A000007, A000142, A240172, A276893, A276894, A276895, A276896, A276897, A276898, A276899, A276900.
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KEYWORD
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AUTHOR
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STATUS
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approved
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