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Triangle T(n,k) = number of permutations of n elements with k 2-cycles.
6

%I #21 Aug 17 2023 18:52:12

%S 1,1,1,1,3,3,15,6,3,75,30,15,435,225,45,15,3045,1575,315,105,24465,

%T 12180,3150,420,105,220185,109620,28350,3780,945,2200905,1100925,

%U 274050,47250,4725,945,24209955,12110175,3014550,519750,51975,10395,290529855

%N Triangle T(n,k) = number of permutations of n elements with k 2-cycles.

%C 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

%H Alois P. Heinz, <a href="/A114320/b114320.txt">Rows n = 0..200, flattened</a>

%F 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).

%F T(n,k) = n!/(2^k*k!) * Sum_{j=0..floor(n/2)-k} (-1/2)^j/j!. - _Alois P. Heinz_, Nov 30 2011

%e T(3,1) = 3 because we have (1)(23), (12)(3) and (13)(2).

%e Triangle begins:

%e 1;

%e 1;

%e 1, 1;

%e 3, 3;

%e 15, 6, 3;

%e 75, 30, 15;

%e 435, 225, 45, 15;

%e ...

%p 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

%t 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 *)

%Y Cf. A008290, A000266, A000090, A088436, A000138, A060725, A060726, A086659, A105422, A105114.

%K easy,nonn,tabf

%O 0,5

%A _Vladeta Jovovic_, Feb 05 2006

%E More terms from _Emeric Deutsch_, Feb 17 2006