OFFSET
2,2
COMMENTS
Determinant of the k-by-k matrix with (i,j)-entry L((i+j)/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 3 < p_n == 3 (mod 4), then a(n) = 0.
It appears that a(n) is even, if p_n == 1 (mod 4).
For any odd prime p, (p+1)/2-i+(p+1)/2-j == -(i+j-1) (mod p) and hence we have L(-1/p)*|L((i+j)/p)|_{i,j=1,...,(p-1)/2} = |L((i+j-1)/p)|_{i,j=1,...,(p-1)/2}. Thus the value of a(n) was actually determined in the first reference of R. Chapman. - Zhi-Wei Sun, Aug 21 2013
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_4 = 7 = 2*3 + 1 and the 3 X 3 matrix (L((i+j)/7)) is
1, -1, 1
-1, 1, -1
1, -1, -1
which has determinant 0, so a(4) = 0.
MATHEMATICA
a[n_] := Module[{p, k}, p = Prime[n]; k = (p-1)/2; Det @ Table[JacobiSymbol[ i + j, p], {i, 1, k}, {j, 1, k}]];
Table[a[n], {n, 2, 32}] (* Jean-François Alcover, Nov 18 2018 *)
CROSSREFS
KEYWORD
sign
AUTHOR
Jonathan Sondow and Wadim Zudilin, Jun 29 2010
STATUS
approved