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A145877
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Triangle read by rows: T(n,k) is the number of permutations of [n] for which the shortest cycle length is k (1<=k<=n).
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14
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1, 1, 1, 4, 0, 2, 15, 3, 0, 6, 76, 20, 0, 0, 24, 455, 105, 40, 0, 0, 120, 3186, 714, 420, 0, 0, 0, 720, 25487, 5845, 2688, 1260, 0, 0, 0, 5040, 229384, 52632, 22400, 18144, 0, 0, 0, 0, 40320, 2293839, 525105, 223200, 151200, 72576, 0, 0, 0, 0, 362880, 25232230
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OFFSET
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1,4
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COMMENTS
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Row sums are the factorials (A000142).
Sum(T(n,k), k=2..n) = A000166(n) (the derangement numbers).
For the statistic "length of the longest cycle", see A126074.
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LINKS
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FORMULA
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E.g.f. for column k is (1-exp(-x^k/k))*exp( -sum(j=1..k-1, x^j/j ) ) / (1-x). - Vladeta Jovovic
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EXAMPLE
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T(4,2)=3 because we have 3412=(13)(24), 2143=(12)(34) and 4321=(14)(23).
Triangle starts:
1;
1, 1;
4, 0, 2;
15, 3, 0, 6;
76, 20, 0, 0, 24;
455, 105, 40, 0, 0, 120;
3186, 714, 420, 0, 0, 0, 720;
25487, 5845, 2688, 1260, 0, 0, 0, 5040;
...
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MAPLE
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F:=proc(k) options operator, arrow: (1-exp(-x^k/k))*exp(-(sum(x^j/j, j = 1 .. k-1)))/(1-x) end proc: for k to 16 do g[k]:= series(F(k), x=0, 16) end do: T:= proc(n, k) options operator, arrow: factorial(n)*coeff(g[k], x, n) end proc: for n to 11 do seq(T(n, k), k=1..n) end do; # yields sequence in triangular form
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MATHEMATICA
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Rest[Transpose[Table[Range[0, 16]! CoefficientList[
Series[(Exp[x^n/n] -1) (Exp[-Sum[x^k/k, {k, 1, n}]]/(1 - x)), {x, 0, 16}], x], {n, 1, 8}]]] // Grid (* Geoffrey Critzer, Mar 04 2011 *)
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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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