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%I
%S 24,240,2240,21840,228480,2580480,31449600,412473600,5801241600,
%T 87178291200,1394852659200,23683435776000,425430061056000,
%U 8062248370176000,160770717499392000,3365514444644352000,73798027581358080000,1691677863018823680000,40464026199993876480000
%N Number of n node tournaments that have exactly two circular triads.
%H J. B. Kadane, <a href="https://doi.org/10.1214/aoms/1177699532">Some equivalence classes in paired comparisons</a>, The Annals of Mathematical Statistics, 37 (1966), 488-494.
%F a(n) = n!*(n - 3 + (1/18)*(n - 4)*(n - 5)) (proven by Kadane).
%e For n = 4 the a(4) = 24 solution is 4!*(4 - 3 + (1/18)*(4 - 4)*(4 - 5)) = 24.
%o (R) fact(n)*(n-3+(1/18)*(n-4)*(n-5))
%K nonn,easy
%O 4,1
%A _Ian R Harris_, Sep 19 2022
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