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A179071 Chapman's "evil" determinants I. 6

%I #27 Aug 25 2021 12:50:57

%S 1,-2,1,1,-18,-4,1,1,-70,1,-882,-32,1,1,-182,1,-29718,1,1,-1068,1,1,

%T -500,-5604,-4030,1,1,-8890182,-776,1,1,-1744,1,-113582,1,-4832118,1,

%U 1,-1118,1,-1111225770,1,-1764132,-11018,1,1,1,1,-20000849130,-23156,1

%N Chapman's "evil" determinants I.

%C 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.

%C 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.)

%C 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.)

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

%H Robin Chapman, <a href="http://dx.doi.org/10.4064/aa115-3-4">Determinants of Legendre symbol matrices</a>, Acta Arith. 115 (2004), 231-244.

%H Robin Chapman, <a href="http://dx.doi.org/10.1016/j.ejc.2003.02.001">Steinitz classes of unimodular lattices</a>, European J. Combin. 25 (2004), 487-493.

%H Robin Chapman (2009), <a href="http://secamlocal.ex.ac.uk/people/staff/rjchapma/etc/evildet.pdf"> My evil determinant problem</a>

%H Maxim Vsemirnov (2011), <a href="http://arxiv.org/abs/1108.4031">On R. Chapman's ``evil determinant'': case p=1 (mod 4)</a>, arXiv:1108.4031 [math.NT], 2011-2012.

%H M. Vseminov, <a href="http://dx.doi.org/10.1016/j.laa.2011.08.039">On the evaluation of R. Chapman's "evil determinant"</a>, Linear Algebra Appl. 436(2012), 4101-4106.

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Legendre_symbol"> Legendre symbol</a>

%e p_3 = 5 = 2*2+1 and the (2+1)-by-(2+1) matrix (L((j-i)/5)) is

%e 0, 1, -1

%e 1, 0, 1

%e -1, 1, 0

%e which has determinant -2, so a(3) = -2.

%t 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}]];

%t Table[a[n], {n, 2, 52}] (* _Jean-François Alcover_, Nov 18 2018 *)

%o (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

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

%K sign

%O 2,2

%A _Jonathan Sondow_ and _Wadim Zudilin_, Jun 29 2010

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Last modified April 19 16:52 EDT 2024. Contains 371794 sequences. (Running on oeis4.)