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%I #48 Oct 13 2022 12:45:36
%S 280,6240,75600,954240,12579840,175392000,2594592000,40721049600,
%T 677053977600,11901451161600,220690229760000,4307253350400000,
%U 88289523818496000,1896762491559936000,42625344258072576000,1000193047805952000000,24463730767033958400000,622724156293184225280000
%N Number of n-node tournaments that have exactly four 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!*((7/3)*(n-4)+4*(n-5)+(7/6)(n-6)(n-7)[n>5]+(1/18)*(n-7)(n-8)(n-9)[n>6]+(1/1944)[n>7]*(n-8)!/(n-12)!) (see Kadane).
%F E.g.f.: (x^7-27*x^6+216*x^5-702*x^4+972*x^3-405*x^2-243*x+189)*x^5/((3^4)*(1-x)^5).
%e For n=5, the a(5)=280 solution is 5!*((7/3)*(5-4)+4*(5-5)+(7/6)(5-6)(5-7)[5>5]+(1/18)*(5-7)(5-8)(5-9)[5>6]+(1/1944)[5>7]*(5-8)!/(5-12)!)=5!*(7/3)*(5-4)=280.
%t CoefficientList[Series[(x^7-27*x^6+216*x^5-702*x^4+972*x^3-405*x^2-243*x+189)*x^5/((3^4)*(1-x)^5), {x,0,22}], x]Table[n!, {n,0,22}] (* _Stefano Spezia_, Sep 27 2022 *)
%Y Cf. A357242, A357257, A357266.
%K nonn
%O 5,1
%A _Ian R Harris_, _Ryan P. A. McShane_, Sep 22 2022
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