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

%I A277692 #49 Mar 23 2025 17:37:48
%S A277692 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,
%T A277692 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,
%U A277692 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
%N A277692 Mendelsohn-Rodney sequence: number of court balanced tournament designs that are available for a given set of teams n.
%C A277692 From _Bernard Schott_, Sep 22 2022: (Start)
%C A277692 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.
%C A277692 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).
%D A277692 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.
%H A277692 Chai Wah Wu, <a href="/A277692/b277692.txt">Table of n, a(n) for n = 1..10000</a>
%H A277692 The IMO Compendium, <a href="https://imomath.com/othercomp/I/Imo1989.pdf">Problem 1</a>, 30th IMO 1989.
%H A277692 E. Mendelsohn and P. Rodney, <a href="http://dx.doi.org/10.1016/0012-365X(94)90027-2">The existence of court balanced tournament designs</a>, Discrete Mathematics, 133 (1994), 207-216.
%H A277692 <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.
%F A277692 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).
%F A277692 a(2n) = A000005(2n-1)-1 for n > 1. - _Chai Wah Wu_, Oct 29 2016
%F A277692 a(1) = 0; a(2n+1) = A183063(2n) for n > 0 - _Bernard Schott_, Sep 22 2022
%e A277692 For n = 9, c = 1, 2, 4 satisfy the conditions given in the formula, so a(n) = 3.
%t A277692 {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 *)
%o A277692 (Python)
%o A277692 import scipy.stats
%o A277692 teams = 151
%o A277692 courtcount = []
%o A277692 i= 1
%o A277692 for j in range(1,teams):
%o A277692     t = 0
%o A277692     for i in range(1,int(j/2) + 1):
%o A277692         if j>1 and ((j*(j-1))/2)%i == 0 and (j-1)%i == 0:
%o A277692             t += 1
%o A277692     if j > 1:
%o A277692         courtcount.append(t)
%o A277692 a = 0
%o A277692 for p in courtcount:
%o A277692     if p == 1:
%o A277692         a+=1
%o A277692 print(courtcount)
%o A277692 (PARI) a(n) = #select(x->(!(binomial(n, 2) % x)) && !((n-1) % x), vector(n\2, k, k)); \\ _Michel Marcus_, Oct 27 2016
%o A277692 (Python)
%o A277692 from __future__ import division
%o A277692 from sympy import divisors
%o A277692 def A277692(n):
%o A277692     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
%Y A277692 Cf. A000005, A183063, A274332.
%K A277692 nonn
%O A277692 1,5
%A A277692 _Andrew G. McEachern_, Oct 27 2016