A102781 -id:A102781 - cp's OEIS frontend

A274332 Team size n for which there exists a balanced tournament for 2n+1 players so that in 2n+1 matches each player plays exactly n-1 times with and n times against each other player.

1, 2, 3, 5, 6, 8, 9, 11, 14, 15, 18, 20, 21, 23, 26, 29, 30, 33

Offset: 0

Michael Steyer, Jun 22 2016

There are 2n+1 players and 2n+1 matches. In each match one person rests, and the remaining 2n players are divided into two equal teams.
Up to n=33 there is probably only a unique design (up to permutation), and it has point / mirror symmetry.
It is conjectured that this sequence is identical to A005097 (ref. Kohen link).
a(n) = A130290(n+2) = A102781(n+2) = A139791(n+1) = A005097(n+1) for 0 <= n <= 17. - Georg Fischer, Oct 30 2018

			n=5:
Match 1: 1,2,3,5,8 versus 4,7,9,10,11
Match 2: 2,3,4,6,9 versus 5,8,10,11,1
Matches 3..11: further cyclic permutations

Daniel Kohen and Ivan Sadofschi, A New Approach on the Seating Couples Problem, arXiv:1006.2571 [math.CO], 2010.

Cf. A005097, A102781, A130290, A139791.

Conjectured design scheme for Team 1 (N:= 2n+1; here players count from 0..2n): X, X+1 (mod N), X+1+2 (mod N), X+1+2+3 (mod N), ...; X = 0..2n (match number). Resting player: (X + (n*(n+1)/2) (mod N).

A371124 a(n) is the least nonnegative integer y such that y^2 = x^2 - k*n for k and x where n > k >= 1 and n > x >= floor(sqrt(n)).

Original entry on oeis.org

0, 0, 1, 0, 2, 2, 3, 1, 0, 4, 5, 2, 6, 6, 1, 0, 8, 0, 9, 4, 2, 10, 11, 1, 0, 12, 3, 6, 14, 2, 15, 2, 4, 16, 1, 0, 18, 18, 5, 3, 20, 4, 21, 10, 2, 22, 23, 1, 0, 0, 7, 12, 26, 6, 3, 5, 8, 28, 29, 2, 30, 30, 1, 0, 4, 8, 33, 16, 10, 2, 35, 3, 36, 36, 5, 18, 2, 10

Offset: 1

Darío Clavijo, Mar 11 2024

nonn

a(A000290(n)) = 0.
a(A077591(n)) = 0.
a(A005563(n)) = 1.
For each n: k = A138191(n) and x = A306284(n).

			 n  | k | x | y^2 = x^2 - k*n  | y
------------------------------------
 1  | 1 | 1 | 0^2 = 1^2 - 1*1  | 0
 2  | 2 | 2 | 0^2 = 2^2 - 2*1  | 0
 11 | 1 | 6 | 5^2 = 6^2 - 1*11 | 5

Cf. A000040, A000290, A005563, A077591, A102781, A138191, A306284.

from sympy.core.power import isqrt
from sympy.ntheory.primetest import is_square
def a(n):
  x = isqrt(n)
  while True:
    for y2 in range(x**2-n, -1, -n):
      if is_square(y2): return isqrt(y2)
    x+=1
print([a(n) for n in range(1, 79)])

from itertools import count
def A371124(n):
    y, a = 0, {}
    for x in count(0):
        if y in a: return a[y]
        a[y] = x
        y = (y+(x<<1)+1)%n # Chai Wah Wu, Apr 25 2024

a(n) = floor(sqrt(A306284(n)^2 - n*A138191(n))).

a(A000040(n)) = A102781(n).

A385326 The number of positive k <= 2n + 1 such that 2n + 1 divides (2^k + 2*n + 1)^2 - 1.

Original entry on oeis.org

1, 3, 2, 2, 3, 2, 2, 7, 4, 2, 7, 2, 2, 3, 2, 6, 6, 5, 2, 6, 4, 6, 7, 2, 2, 12, 2, 5, 6, 2, 2, 21, 10, 2, 6, 2, 8, 7, 5, 2, 3, 2, 21, 6, 8, 15, 18, 5, 4, 6, 2, 2, 17, 2, 6, 6, 8, 5, 19, 9, 2, 12, 2, 18, 18, 2, 14, 7, 4, 2, 6, 4, 10, 7, 2, 10, 12, 15, 6, 6, 4, 2, 16, 2, 2, 19, 2, 5, 6, 2, 2, 6, 10, 9, 21, 2, 4, 32, 2, 2, 6

Offset: 0

Juri-Stepan Gerasimov, Jun 25 2025

nonn

			1 is the term because 2*0 + 1 = 1 is divisor of (2^1 + 2*0 + 1)^2 - 1 = 3^2 - 1 = 8.

Cf. A003462 (numbers m > 0 such that a(m) = 3), A005384 (primes p such that a(p) = 2), A005408 (odd numbers), A076481 (primes q such that a(q) = 3), A081858 (numbers k numbers k >= 0 such that 2k + 1 divides 2^k - 1), A102781 (numbers k such that 2k + 1 divides (2^k + 2*k + 1)^2 - 1), A224486 (numbers k such that 2k + 1 divides 2^k + 1).

[#[k: k in [1..2*n+1] | ((2^k+2*n+1)^2 - 1) mod (2*n + 1) eq 0]: n in [0..100]];

a[n_]:=Length[Select[Range[2n+1],Divisible[(2^#+2n+1)^2-1,2n+1] &]]; Array[a,101,0] (* Stefano Spezia, Jun 25 2025 *)

a(n) = sum(k=1, 2*n+1, !Mod((2^k + 2*n + 1)^2 - 1, 2*n + 1)); \\ Michel Marcus, Jun 25 2025

cp's OEIS Frontend

A274332 Team size n for which there exists a balanced tournament for 2n+1 players so that in 2n+1 matches each player plays exactly n-1 times with and n times against each other player.

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A371124 a(n) is the least nonnegative integer y such that y^2 = x^2 - k*n for k and x where n > k >= 1 and n > x >= floor(sqrt(n)).

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A385326 The number of positive k <= 2n + 1 such that 2n + 1 divides (2^k + 2*n + 1)^2 - 1.

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Magma

Mathematica

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cp's OEIS Frontend

A274332 Team size n for which there exists a balanced tournament for 2n+1 players so that in 2n+1 matches each player plays exactly n-1 times with and n times against each other player.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

A371124 a(n) is the least nonnegative integer y such that y^2 = x^2 - k*n for k and x where n > k >= 1 and n > x >= floor(sqrt(n)).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Python

Python

Formula

A385326 The number of positive k <= 2*n + 1 such that 2*n + 1 divides (2^k + 2*n + 1)^2 - 1.

Views

Author

Keywords

Examples

Crossrefs

Programs

Magma

Mathematica

PARI

A385326 The number of positive k <= 2n + 1 such that 2n + 1 divides (2^k + 2*n + 1)^2 - 1.