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A114320
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Triangle T(n,k) = number of permutations of n elements with k 2-cycles.
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6
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1, 1, 1, 1, 3, 3, 15, 6, 3, 75, 30, 15, 435, 225, 45, 15, 3045, 1575, 315, 105, 24465, 12180, 3150, 420, 105, 220185, 109620, 28350, 3780, 945, 2200905, 1100925, 274050, 47250, 4725, 945, 24209955, 12110175, 3014550, 519750, 51975, 10395, 290529855
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OFFSET
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0,5
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COMMENTS
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Row n has 1+floor(n/2) terms. Row sums yield the factorials (A000142). Sum(k*T(n,k),k>0)=n!/2 for n>=2. - Emeric Deutsch, Feb 17 2006
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LINKS
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FORMULA
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E.g.f.: exp((y-1)*x^2/2)/(1-x). More generally, e.g.f. for number of permutations of n elements with k m-cycles is exp((y-1)*x^m/m)/(1-x).
T(n,k) = n!/(2^k*k!) * Sum_{j=0..floor(n/2)-k} (-1/2)^j/j!. - Alois P. Heinz, Nov 30 2011
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EXAMPLE
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T(3,1) = 3 because we have (1)(23), (12)(3) and (13)(2).
Triangle begins:
1;
1;
1, 1;
3, 3;
15, 6, 3;
75, 30, 15;
435, 225, 45, 15;
...
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MAPLE
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G:= exp((y-1)*x^2/2)/(1-x): Gser:= simplify(series(G, x=0, 15)): P[0]:=1: for n from 1 to 12 do P[n]:= n!*coeff(Gser, x^n) od: for n from 0 to 12 do seq(coeff(y*P[n], y^j), j=1..1+floor(n/2)) od; # yields sequence in triangular form - Emeric Deutsch, Feb 17 2006
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MATHEMATICA
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d = Exp[-x^2/2!]/(1 - x); f[list_] := Select[list, # > 0 &]; Flatten[Map[f, Transpose[Table[Range[0, 10]!CoefficientList[Series[x^(2 k)/(2^k k!) d, {x, 0, 10}], x], {k, 0, 5}]]]] (* Geoffrey Critzer, Nov 29 2011 *)
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CROSSREFS
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KEYWORD
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easy,nonn,tabf
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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