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.

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

1,6

Links

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.

Eric Weisstein's World of Mathematics, Perfect Difference Set

Example

   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.

Prog

(PARI) 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)); );

Crossrefs

Cf. A002061, A058241, A351690, A333852.

Sequence in context: A248949 A325673 A083857 * A115142 A346203 A327126

Adjacent sequences: A353074 A353075 A353076 * A353078 A353079 A353080

Keyword

sign,look,tabl

Author

Michel Marcus, Apr 22 2022

Extensions

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

Status

approved