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

280, 6240, 75600, 954240, 12579840, 175392000, 2594592000, 40721049600, 677053977600, 11901451161600, 220690229760000, 4307253350400000, 88289523818496000, 1896762491559936000, 42625344258072576000, 1000193047805952000000, 24463730767033958400000, 622724156293184225280000

Offset

5,1

Links

Table of n, a(n) for n=5..22.

Ian R. Harris and Ryan P. A. McShane, Counting Tournaments with a Specified Number of Circular Triads, Journal of Integer Sequences, Vol. 27 (2024), Article 24.8.7. See pages 2, 23.

J. B. Kadane, Some equivalence classes in paired comparisons, The Annals of Mathematical Statistics, 37 (1966), 488-494.

Formula

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).

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).

Example

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.

Mathematica

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 *)

Crossrefs

Cf. A357242, A357257, A357266.

Sequence in context: A187384 A297724 A024214 * A094895 A223107 A218411

Adjacent sequences: A357245 A357246 A357247 * A357249 A357250 A357251

Keyword

nonn

Author

Ian R Harris, Ryan P. A. McShane, Sep 22 2022

Status

approved