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

1,2

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).

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

Example

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.

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 *)

Prog

(PARI) for(n=1, 30, print1((2/5!)*n*(n-1)*(n+3)*(n+2)*(n+1), ", ")) \\ G. C. Greubel, Nov 25 2017
(Magma) [(2/Factorial(5))*n*(n-1)*(n+3)*(n+2)*(n+1): n in [1..30]]; // G. C. Greubel, Nov 25 2017

Crossrefs

Cf. A000389.

Sequence in context: A364598 A094702 A001621 * A290928 A232584 A189491

Adjacent sequences: A277932 A277933 A277934 * A277936 A277937 A277938

Keyword

nonn,easy

Author

Andrew Lohr, Nov 04 2016

Status

approved