A179071 Chapman's "evil" determinants I.

1, -2, 1, 1, -18, -4, 1, 1, -70, 1, -882, -32, 1, 1, -182, 1, -29718, 1, 1, -1068, 1, 1, -500, -5604, -4030, 1, 1, -8890182, -776, 1, 1, -1744, 1, -113582, 1, -4832118, 1, 1, -1118, 1, -1111225770, 1, -1764132, -11018, 1, 1, 1, 1, -20000849130, -23156, 1

Offset: 2

Author

Jonathan Sondow and Wadim Zudilin, Jun 29 2010

Keywords

• sign

Comments

Determinant of the (k+1)X(k+1) matrix with (i,j)-entry L((j-i)/p), where L(./p) denotes the Legendre symbol modulo p and p = p_n = 2k+1 is the n-th prime.

Guy says "Chapman has a number of conjectures which concern the distribution of quadratic residues." One is that if p_n == 3 (mod 4), then a(n) = 1. Chapman also has a conjecture if p_n == 1 (mod 4), involving the fundamental unit and class number of the quadratic field Q(sqrt(p)). (Added Aug 23 2011: Both conjectures have been proved by Vsemirnov.)

It appears that a(n) is negative and even, if p_n == 1 (mod 4); see A179073. (Added Aug 28 2011: This conjecture has also been proved by Vsemirnov.)

Examples

			p_3 = 5 = 2*2+1 and the (2+1)-by-(2+1) matrix (L((j-i)/5)) is
   0,  1, -1
   1,  0,  1
  -1,  1,  0
which has determinant -2, so a(3) = -2.
		

References

• Richard Guy, Unsolved Problems in Number Theory, 3rd ed., Springer, 2004, Section F5.

Links

• Robin Chapman, Determinants of Legendre symbol matrices, Acta Arith. 115 (2004), 231-244.
• Robin Chapman, Steinitz classes of unimodular lattices, European J. Combin. 25 (2004), 487-493.
• Robin Chapman, My evil determinant problem, 2009.
• Maxim Vsemirnov, On R. Chapman's "evil determinant": case p=1 (mod 4), arXiv:1108.4031 [math.NT], 2011-2012.
• Maxim Vsemirnov, On the evaluation of R. Chapman's "evil determinant", Linear Algebra Appl. 436(2012), 4101-4106.
• Wikipedia, Legendre symbol

Crossrefs

Cf. A179072 (Chapman's "evil" determinants II), A179073 (A179071 for p == 1 (mod 4)), A179074 (A179072 for p == 1 (mod 4)).

Programs

Mathematica

a[n_] := Module[{p, k}, p = Prime[n]; k = (p-1)/2; Det @ Table[JacobiSymbol[ j-i, p], {i, 1, k+1}, {j, 1, k+1}]];
Table[a[n], {n, 2, 52}] (* Jean-François Alcover, Nov 18 2018 *)

PARI

a(n) = my(p=prime(n), k=(p+1)/2); matdet(matrix(k, k, i, j, kronecker(j-i, p))); \\ Michel Marcus, Aug 25 2021