A277935 Number of ways 2*n-1 people can vote on three candidates so that the Condorcet paradox arises.

0, 2, 12, 42, 112, 252, 504, 924, 1584, 2574, 4004, 6006, 8736, 12376, 17136, 23256, 31008, 40698, 52668, 67298, 85008, 106260, 131560, 161460, 196560, 237510, 285012, 339822, 402752, 474672, 556512, 649264, 753984, 871794, 1003884, 1151514, 1316016, 1498796

Offset: 1

Author

Andrew Lohr, Nov 04 2016

Keywords

• nonn
• easy

Examples

			For n=2 (three voters), the two possible ways the Condorcet paradox arises are:
1) one voter prefers A to B to C, one prefers B to C to A, and one prefers C to A to B.
2) one voter prefers A to C to B, one prefers C to B to A, and one prefers B to A to C.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000
• R. Embar, D. Zeilberger,Counting Condorcet, Enum. Combin. Applic. 2 (2022) #S2R22
• Wikipedia, Condorcet paradox
• Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Crossrefs

Cf. A000389.

Programs

Magma

[(2/Factorial(5))*n*(n-1)*(n+3)*(n+2)*(n+1): n in [1..30]]; // G. C. Greubel, Nov 25 2017

Mathematica

Table[(2/5!)*n*(n - 1)*(n + 3)*(n + 2)*(n + 1), {n, 1, 50}] (* G. C. Greubel, Nov 25 2017 *)
a[n_] := 2 Binomial[n + 3, 5]; Array[a, 40] (* or *)
Rest@ CoefficientList[ Series[2 x^2/(x - 1)^6, {x, 0, 40}], x] (* or *)
Range[0, 40]! CoefficientList[ Series[x^2 (x^3 + 15x^2 + 60x + 60) Exp[x]/60, {x, 0, 40}], x] (* or *)
LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 2, 12, 42, 112, 252, 504}, 40] (* Robert G. Wilson v, Nov 25 2017 *)

PARI

for(n=1,30, print1((2/5!)*n*(n-1)*(n+3)*(n+2)*(n+1), ", ")) \\ G. C. Greubel, Nov 25 2017

Formula

a(n) = (2/5!)*n*(n-1)*(n+3)*(n+2)*(n+1).

From N. J. A. Sloane, Nov 10 2016: (Start)

a(n) = 2*binomial(n+3,5) = 2*A000389(n+3).

G.f.: 2*x^2/(1-x)^6. (End)

E.g.f.: x^2*(60 + 60*x + 15*x^2 + x^3)*exp(x)/60. - G. C. Greubel, Nov 25 2017