OFFSET
2,2
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.)
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 (2009), My evil determinant problem
Maxim Vsemirnov (2011), On R. Chapman's ``evil determinant'': case p=1 (mod 4), arXiv:1108.4031 [math.NT], 2011-2012.
M. Vseminov, On the evaluation of R. Chapman's "evil determinant", Linear Algebra Appl. 436(2012), 4101-4106.
Wikipedia, Legendre symbol
EXAMPLE
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.
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 *)
PROG
(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
CROSSREFS
KEYWORD
sign
AUTHOR
Jonathan Sondow and Wadim Zudilin, Jun 29 2010
STATUS
approved