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Search: id:A161123
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| A161123 |
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Triangle read by rows: T(n,k) is the number of fixed-point-free involutions of {1,2,...,2n} having k inversions (0<=k<=n(2n-1)). |
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+0
2
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| 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 2, 0, 3, 0, 3, 0, 3, 0, 2, 0, 1, 0, 0, 0, 0, 1, 0, 3, 0, 6, 0, 9, 0, 12, 0, 14, 0, 15, 0, 14, 0, 12, 0, 9, 0, 6, 0, 3, 0, 1, 0, 0, 0, 0, 0, 1, 0, 4, 0, 10, 0, 19, 0, 31, 0, 45, 0, 60, 0, 74, 0, 86, 0, 94, 0, 97, 0, 94, 0, 86, 0, 74, 0, 60, 0, 45, 0, 31, 0, 19
(list; graph; listen)
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OFFSET
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0,16
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COMMENT
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Sum of entries in row n is (2n-1)!! =A001147(n).
Row n has 1+2n(n-1) entries.
Sum(k*T(n,k), k>=0) = (2n-1)!!*n^2 = A161124(n).
A128080 is the same triangle with the 0's deleted.
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FORMULA
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Generating polynomial of row n is P_n(q)=[q/(1-q^2)]^n*Product(1-q^{4j-2}, j=1..n).
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EXAMPLE
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T(3,11)=3 because we have 465132, 546213, and 632541.
Triangle starts:
1;
0,1;
0,0,1,0,1,0,1;
0,0,0,1,0,2,0,3,0,3,0,3,0,2,0,1.
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MAPLE
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f := proc (n) options operator, arrow: q^n*(product(1-q^(4*j-2), j = 1 .. n))/(1-q^2)^n end proc: for n from 0 to 4 do P[n] := sort(expand(simplify(f(n)))) end do: for n from 0 to 4 do seq(coeff(P[n], q, j), j = 0 .. n*(2*n-1)) end do; # yields sequence in triangular form
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CROSSREFS
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A001147, A161124, A128080
Sequence in context: A127460 A154109 A011374 this_sequence A035442 A035376 A029220
Adjacent sequences: A161120 A161121 A161122 this_sequence A161124 A161125 A161126
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KEYWORD
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nonn,tabf
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AUTHOR
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Emeric Deutsch (deutsch(AT)duke.poly.edu), Jun 05 2009
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