A274332 -id:A274332 - cp's OEIS frontend

A277692 Mendelsohn-Rodney sequence: number of court balanced tournament designs that are available for a given set of teams n.

0, 1, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 4, 1, 2, 3, 4, 1, 3, 1, 4, 3, 2, 1, 6, 2, 2, 3, 4, 1, 4, 1, 5, 3, 2, 3, 6, 1, 2, 3, 6, 1, 4, 1, 4, 5, 2, 1, 8, 2, 3, 3, 4, 1, 4, 3, 6, 3, 2, 1, 8, 1, 2, 5, 6, 3, 4, 1, 4, 3, 4, 1, 9, 1, 2, 5, 4, 3, 4, 1, 8, 4, 2, 1, 8, 3, 2, 3, 6, 1, 6, 3, 4, 3, 2, 3, 10, 1, 3, 5, 6, 1, 4, 1, 6, 7, 2, 1

Offset: 1

Andrew G. McEachern, Oct 27 2016

nonn

From Bernard Schott, Sep 22 2022: (Start)
a(n+1) is the number of solutions (n, m) such that the set An = {1, 2, ..., n} can be expressed as the disjoint union of m subsets E1, E2, ..., Em that satisfy: (i) each Ei contains the same number of elements, and, (ii) the sum of elements of each Ei is the same for i = 1, 2, ..., m.
The 1st problem, proposed by Philippines, during the 30th International Mathematical Olympiad in 1989 at Braunschweig - Niedersachen (Brunswick), FR Germany, asked to prove there is a solution for n = 1989 and m = 117 (see Holton and IMO link) (End).

			For n = 9, c = 1, 2, 4 satisfy the conditions given in the formula, so a(n) = 3.

Derek Holton, A First Step to Mathematical Olympiad Problems, Vol. 1, Mathematical Olympiad Series, World Scientific, 2010, & 8.3. PHIL 1 pp. 250-252 and & 8.11 Solutions pp 261-265.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
The IMO Compendium, Problem 1, 30th IMO 1989.
E. Mendelsohn and P. Rodney, The existence of court balanced tournament designs, Discrete Mathematics, 133 (1994), 207-216.
Index to sequences related to Olympiads.

Cf. A000005, A183063, A274332.

{0}~Join~Table[Function[k, Count[Range@ k, c_ /; And[Divisible[n - 1, c], Divisible[Binomial[n, 2], c]] ]]@ Floor[n/2], {n, 2, 108}] (* Michael De Vlieger, Oct 27 2016 *)

a(n) = #select(x->(!(binomial(n, 2) % x)) && !((n-1) % x), vector(n\2, k, k)); \\ Michel Marcus, Oct 27 2016

import scipy.stats
teams = 151
courtcount = []
i= 1
for j in range(1,teams):
    t = 0
    for i in range(1,int(j/2) + 1):
        if j>1 and ((j*(j-1))/2)%i == 0 and (j-1)%i == 0:
            t += 1
    if j > 1:
        courtcount.append(t)
a = 0
for p in courtcount:
    if p == 1:
        a+=1
print(courtcount)

from _future_ import division
from sympy import divisors
def A277692(n):
    return sum(1 for c in divisors(n-1) if c < n-1 and not (n*(n-1)//2) % c) if n != 2 else 1 # Chai Wah Wu, Oct 29 2016

a(n) is the number of values of c that satisfy the following conditions: C(n,2) mod c = 0, and (n-1) mod c = 0, and 1 <= c <= floor(n/2).
a(2n) = A000005(2n-1)-1 for n > 1. - Chai Wah Wu, Oct 29 2016
a(1) = 0; a(2n+1) = A183063(2n) for n > 0 - Bernard Schott, Sep 22 2022

cp's OEIS Frontend

A277692 Mendelsohn-Rodney sequence: number of court balanced tournament designs that are available for a given set of teams n.

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