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A357248 Number of n-node tournaments that have exactly four circular triads.

%I A357248 #52 Jan 06 2025 06:31:19
%S A357248 280,6240,75600,954240,12579840,175392000,2594592000,40721049600,
%T A357248 677053977600,11901451161600,220690229760000,4307253350400000,
%U A357248 88289523818496000,1896762491559936000,42625344258072576000,1000193047805952000000,24463730767033958400000,622724156293184225280000
%N A357248 Number of n-node tournaments that have exactly four circular triads.
%H A357248 Ian R. Harris and Ryan P. A. McShane, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL27/McShane/mcshane1.html">Counting Tournaments with a Specified Number of Circular Triads</a>, Journal of Integer Sequences, Vol. 27 (2024), Article 24.8.7. See pages 2, 23.
%H A357248 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 A357248 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 A357248 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 A357248 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 A357248 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 A357248 Cf. A357242, A357257, A357266.
%K A357248 nonn
%O A357248 5,1
%A A357248 _Ian R Harris_, _Ryan P. A. McShane_, Sep 22 2022