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%I #20 Jan 21 2023 09:08:00
%S 1,1,3,1,18,9,163,90,3,1950,1100,75,28821,16245,1575,15,505876,283122,
%T 33810,735,10270569,5699932,780150,26460,105,236644092,130267440,
%U 19615932,884520,8505,6098971555,3332614725,538325550,29619450,467775,945
%N Triangular array read by rows: T(n,k) is the number of functions f:{1,2,...,n} -> {1,2,...,n} that have exactly k 2-cycles for n >= 0 and 0 <= k <= floor(n/2).
%C It appears that as n gets large, row n conforms to a Poisson distribution with mean = 1/2. In other words, as n gets large, T(n,k) approaches n^n/(2^k*k!*e^(1/2)).
%F E.g.f.: exp((T(x)^2/2)*(y-1))/(1 - T(x)) where T(x) is the e.g.f. for A000169.
%F Sum_{k=1..floor(n/2)} k * T(n,k) = A081131(n).
%e Triangle begins:
%e 1;
%e 1;
%e 3, 1;
%e 18, 9;
%e 163, 90, 3;
%e 1950, 1100, 75;
%e 28821, 16245, 1575, 15;
%e 505876, 283122, 33810, 735;
%e 10270569, 5699932, 780150, 26460, 105;
%e 236644092, 130267440, 19615932, 884520, 8505;
%e 6098971555, 3332614725, 538325550, 29619450, 467775, 945;
%e ...
%t nn=10;t=Sum[n^(n-1)x^n/n!,{n,1,nn}]; Range[0,nn]! CoefficientList[Series[Exp[t^2/2(y-1)]/(1-t), {x,0,nn}], {x,y}]//Grid
%Y Column k=0 gives A089466.
%Y Cf. A000169, A081131.
%K nonn,tabf
%O 0,3
%A _Geoffrey Critzer_, Dec 24 2012
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