OFFSET
1,2
COMMENTS
Teams play each other twice for a total of M = n*(n-1) matches.
A victory is awarded 3 points, a draw 1 point and a defeat 0 points.
The total number of possible match outcomes is 3^M = A053764(n) and a(n) is how many of them result in all teams finishing with the same points score.
If all matches were randomly assigned a result, the probability that all teams would end up with the same number of points is a(n)/A053764(n), which in a typical league of 18 or 20 teams is very small.
A007080(n) is the number of ways if there are no draws.
LINKS
Ruediger Jehn, Julia brute-force program
RĂ¼diger Jehn, Kester Habermann, and Misha Lavrov, Number of ways that a football league can complete with all teams having the same number of points, arXiv:2503.14509 [math.GM], 2025.
Ruediger Jehn, Julia program based on fast algorithm
EXAMPLE
We denote the vector (r1, r2 ... r_M) with r_i in {0, 1, 3} as a possible sequence of match results. Then a(2) = 3: (0, 0) - both teams lose their home game and have 3 points at the end, (1,1) - both matches end with a draw and both teams have 2 points, (3,3) - both teams win their home game and have 3 points.
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Ruediger Jehn, Jan 27 2025
STATUS
approved
