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%I #23 Dec 15 2021 10:25:01
%S 1,1,4,25,2,224,32,2625,500,38056,8560,40,657433,164150,1960,13178880,
%T 3526656,71680,300585601,84389928,2442720,2240,7683776000,2232672000,
%U 83328000,224000,217534555161,64830707370,2931500880,14907200
%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 3-cycles. n>=0, 0<=k<=floor(n/3).
%C The total number of 3-cycles over all functions on {1,2,...,n} is 2*binomial(n,3)*n^(n-3). So we see that as n gets large the probability that a random function would contain k 3-cycles is a Poisson distribution with mean = 1/3. Generally, the total number of j-cycles over all functions on {1,2,...,n} is (j-1)!*binomial(n,j)*n^(n-j).
%F E.g.f.: exp(T(x)^3/3*(y - 1))/(1-T(x)) where T(x) is the e.g.f. for A000169.
%e 1;
%e 1;
%e 4;
%e 25, 2;
%e 224, 32;
%e 2625, 500;
%e 38056, 8560, 40;
%e 657433, 164150, 1960;
%e 13178880, 3526656, 71680;
%e 300585601, 84389928, 2442720, 2240;
%e ...
%t nn=10;t=Sum[n^(n-1)x^n/n!,{n,1,nn}];Range[0,nn]!CoefficientList[ Series[Exp[t^3/3(y-1)]/(1-t),{x,0,nn}],{x,y}]//Grid
%Y Cf. A185025, A055134, A190314.
%K nonn,tabf
%O 0,3
%A _Geoffrey Critzer_, Dec 25 2012