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

1,5

Comments

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).

References

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.

Links

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.

Formula

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

Example

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

Mathematica

{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 *)

Prog

(Python 3.5)
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)
(PARI) a(n) = #select(x->(!(binomial(n, 2) % x)) && !((n-1) % x), vector(n\2, k, k)); \\ Michel Marcus, Oct 27 2016
(Python)
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

Crossrefs

Cf. A000005, A183063, A274332.

Sequence in context: A335079 A337093 A320266 * A354992 A129138 A353378

Adjacent sequences: A277689 A277690 A277691 * A277693 A277694 A277695

Keyword

nonn

Author

Andrew G. McEachern, Oct 27 2016

Status

approved