A353077
Triangle read by rows, where the n-th row consists of the lexicographically earliest solution for n integers in 0..p-1 whose n*(n-1) differences are congruent to 1..p-1 (mod p), where p=n*(n-1)+1. If no solution exists, the n-th row consists of n -1's.
Original entry on oeis.org
0, 0, 1, 0, 1, 3, 0, 1, 3, 9, 0, 1, 4, 14, 16, 0, 1, 3, 8, 12, 18, -1, -1, -1, -1, -1, -1, -1, 0, 1, 3, 13, 32, 36, 43, 52, 0, 1, 3, 7, 15, 31, 36, 54, 63, 0, 1, 3, 9, 27, 49, 56, 61, 77, 81, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, 0, 1, 3, 12, 20, 34, 38, 81, 88, 94, 104, 109
Offset: 1
n row
1 [0];
2 [0,1];
3 [0,1,3];
4 [0,1,3,9];
5 [0,1,4,14,16];
6 [0,1,3,8,12,18];
7 no solution exists;
8 [0,1,3,13,32,36,43,52];
9 [0,1,3,7,15,31,36,54,63];
10 [0,1,3,9,27,49,56,61,77,81];
11 no solution exists;
12 [0,1,3,12,20,34,38,81,88,94,104,109];
13 no solution exists;
14 [0,1,3,16,23,28,42,76,82,86,119,137,154,175];
15 no solution exists;
16 no solution exists.
- Martin Becker, Rows n = 1..200 of triangle, flattened.
- Leonard E. Dickson, Problem 142, The American Mathematical Monthly, Vol. 14, No. 5 (May, 1907), pp. 107-108.
- Daniel Gordon, On difference sets with small lambda, arXiv:2007.07292 [math.CO], 2020.
- Eric Weisstein's World of Mathematics, Perfect Difference Set
-
isok(n, v) = my(p=n*(n-1)+1); setbinop((x,y)->lift(Mod(x-y, p)), v, v) == [0..p-1];
row(n) = forsubset([n^2-n+1, n], s, my(ds = apply(x->x-1, Vec(s))); if (isok(n, ds), return(ds)););
Name and data corrected for "lexicographically earliest solution" by
Michel Marcus, May 09 2022
Adjusted to a regular triangle, and rows 1, 2, 7, and 10-12 inserted by
Pontus von Brömssen, May 09 2022
A102508
Suppose there are equally spaced chairs around a round table. Then a(n) is the maximal number of chairs for which there exists a seating arrangement of n people around the table such that if a waiter puts two glasses (randomly) on the table in front of two (different) chairs, it is always possible to turn the table so that the two glasses end up in front of two seated persons.
Original entry on oeis.org
1, 3, 7, 13, 21, 31, 39, 57, 73, 91, 95, 133
Offset: 1
Ard Van Moer (ard.van.moer(AT)vub.ac.be), Mar 15 2005
a(5)=21 because if we have 21 chairs, 5 persons can sit down on chairs 1, 4, 5, 10 and 12. 1 == 5-4 (mod 21). 2 == 12-10 (mod 21). 3 == 4-1 (mod 21). 4 == 5-1 (mod 21). 5 == 10-5 (mod 21). 6 == 10-4 (mod 21). 7 == 12-5 (mod 21). 8 == 12-4 (mod 21). 9 == 10-1 (mod 21). 10 == 1-12 (mod 21). It is impossible to do the same with 22 or more chairs.
A351690
a(n) is the number of n-subsets of [0..p-1] whose n*(n-1) differences are congruent to 1..p-1 (mod p), where p=n*(n-1)+1.
Original entry on oeis.org
1, 3, 14, 52, 42, 310, 0, 684, 584, 1092, 0, 4788, 0, 7320, 0, 0, 3276, 31314, 0, 32004, 0, 0, 0
Offset: 1
- Leonard E. Dickson, Problem 142, The American Mathematical Monthly, Vol. 14, No. 5 (May, 1907), pp. 107-108.
- Eric Weisstein's World of Mathematics, Perfect Difference Set
-
isok(n, v) = my(p=n*(n-1)+1); setbinop((x, y)->lift(Mod(x-y, p)), v, v) == [0..p-1];
a(n) = my(nb=0); forsubset([n^2-n+1, n], s, my(ds = apply(x->x-1, Vec(s))); if (isok(n, ds), nb++)); nb;
A353108
a(n) is the number of cycles of n numbers arranged so that every integer in 1..n*(n-1)+1 occurs as the sum of up to n adjacent numbers. Both a solution and its reverse are counted unless they are identical.
Original entry on oeis.org
1, 1, 2, 4, 2, 10, 0, 12, 8, 12, 0, 36, 0, 40, 0, 0, 12, 102, 0, 84, 0, 0, 0
Offset: 1
For n = 1, the only solution consists of the single number { 1 }, and a "cycle" consisting of { 1 } is the same whether read forward or backward, so a(1) = 1.
For n = 2, the only solution (starting at 1) consists of the two numbers { 1, 2 }; arranging these around a circle as
1
/ \
\ /
2
gives the same cycle, i.e., { 1, 2 } whether read clockwise or counterclockwise from 1, so a(2) = 1.
For n = 3, the two cycles (starting at 1) are { 1, 2, 4 } and { 1, 4, 2 }, so a(3) = 2.
For n = 8, the twelve solutions are
{ 1, 2, 10, 19, 4, 7, 9, 5 },
{ 1, 3, 5, 11, 2, 12, 17, 6 },
{ 1, 3, 8, 2, 16, 7, 15, 5 },
{ 1, 4, 2, 10, 18, 3, 11, 8 },
{ 1, 4, 22, 7, 3, 6, 2, 12 },
{ 1, 6, 12, 4, 21, 3, 2, 8 },
and the same six cycles read in the opposite direction from 1 (e.g.,
{ 1, 2, 10, 19, 4, 7, 9, 5 }
read in reverse order starting at 1 is
{ 1, 5, 9, 7, 4, 19, 10, 2 }
each of which counts as a separate solution), so a(8) = 12.
Showing 1-4 of 4 results.
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