A353194 Number of Condorcet voting profiles with three candidates and 2n-1 voters.
0, 12, 540, 21000, 785820, 28956312, 1058809752, 38545567632, 1399354322652, 50707958458872, 1835099465988360, 66348521294296176, 2397139928161319640, 86559958069097395440, 3124302168622853150640, 112729791393354644416800
Offset: 1
Keywords
Links
- Shalosh B. Ekhad, More terms.
- Rebecca Embar and Doron Zeilberger, Counting Condorcet.
- Doron Zeilberger, Condorcet3 Maple package.
Programs
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Maple
#(From Maple package Condorcet3 by Doron Zeilberger) #NuCo(N):The first N terms of the sequence "number of Condorcet vote-profiles" with 2v-1 voters and three candidates. Using the third-order recurrence. Try: #NuCo(100); NuCo:=proc(N) local L,n,kha: L:=[0,12,540]: if N<=3 then RETURN(L[N]): fi: for n from 4 to N do kha:=4*(19*n^2-57*n+45)/(n-1)^2*L[-1]-36*(2*n-3)*(22*n^2-99*n+111)/(n-2)/(n-1)^2*L[-2]+1296*(n-3)*(2*n-3)*(2*n-5)/(n-2)/(n-1)^2*L[-3]: L:=[L[2],L[3],kha]: od: L[-1]: end: seq(NuCo(n), n=1..16);
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Mathematica
RecurrenceTable[{a[n] == (4*(19*n^2 - 57*n + 45)/(n-1)^2) * a[n-1] - (36*(2*n - 3)*(22*n^2 - 99*n + 111)/((n-2)*(n-1)^2)) * a[n-2] + (1296*(n-3)*(2*n - 3)*(2*n - 5)/((n-2)*(n-1)^2)) * a[n-3], a[1] == 0, a[2] == 12, a[3] == 540}, a[n], {n, 1, 20}] (* Vaclav Kotesovec, May 20 2022 *) -
PARI
a(n) = 2*sum(i1=0, n-2, sum(i2=0, n-2-i1, sum(i3=0, n-2-i1-i2, sum(i4=0, n-2-i1-i2-i3, sum(i5=0, n-2-i1-i2-i3-i4, ((2*n-1)!/((n-1-i2-i3-i5)!*i2!*i3!*(i2+i4+i5+1)!*(n-1-i1-i2-i4)!*i1!))))))) \\ Michel Marcus, May 03 2022
Formula
a(n) = (4*(19*n^2-57*n+45)/(n-1)^2)*a(n-1) - (36*(2*n-3)*(22*n^2-99*n+111)/((n-2)*(n-1)^2))*a(n-2) + (1296*(n-3)*(2*n-3)*(2*n-5)/((n-2)*(n-1)^2))*a(n-3).
a(n) = 2*(Sum_{i1=0..n-2} Sum_{i2=0..n-2-i1} Sum_{i3=0..n-2-i1-i2} Sum_{i4=0..n-2-i1-i2-i3} Sum_{i5=0..n-2-i1-i2-i3-i4} ((2*n-1)!/((n-1-i2-i3-i5)!*i2!*i3!*(i2+i4+i5+1)!*(n-1-i1-i2-i4)!*i1!))).
a(n) ~ (1/4 - 3*arcsin(1/3)/(2*Pi)) * 6^(2*n - 1) [Guilbaud, 1952].
Comments