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%I #51 Aug 25 2021 06:56:42
%S 0,-1,0,0,-5,1,0,0,-13,0,-145,5,0,0,-25,0,-3805,0,0,125,0,0,53,569,
%T -401,0,0,-851525,73,0,0,149,0,-9305,0,-385645,0,0,-85,0,-82596761,0,
%U 126985,-785,0,0,0,0,-1321693313,1517,0,4574225,0,1025,0,-134485,0,-535979945,63445,0,-145,0,0,7170685,-19805,0,55335641,0,-167273125693,3793,0,0,-27765559705,0,0,-427316305,-1027776565,2564801,5534176685
%N Determinant of the (p_n-1)/2 X (p_n-1)/2 matrix with (i,j)-entry being the Legendre symbol ((j-i)/p_n), where p_n is the n-th prime.
%C Conjecture: In the case p_n == 1 (mod 4), (2/p_n)*a(n) is a positive odd integer whose prime factors are all congruent to 1 modulo 4, and moreover for some integer b(n) we have b(n) + (2/p_n)*a(n)*sqrt(p_n) = e(p_n)^{(2-(2/p_n))h(p_n)}, where e(p_n) and h(p_n) are the fundamental unit and the class number of the real quadratic field Q(sqrt(p_n)) respectively.
%C Note that a(n) = 0 when p_n == 3 (mod 4), this is because the transpose of the determinant a(n) coincides with (-1/p_n)^{(p_n-1)/2}*a(n) = -a(n).
%C M. Vsemirnov has proved _Robin Chapman_'s conjecture on the evaluation of the determinant of the (p+1)/2-by-(p+1)/2 matrix with (i,j)-entry (i,j = 0,...,(p-1)/2) being the Legendre symbol ((j-i)/p), where p is an odd prime.
%C On Aug 14 2013, _Robin Chapman_ informed the author that he first made the conjecture about the exact value of a(n) in a private manuscript dated Aug 05 2003.
%H Zhi-Wei Sun, <a href="/A228077/b228077.txt">Table of n, a(n) for n = 2..200</a>
%H R. C. Chapman, <a href="http://empslocal.ex.ac.uk/people/staff/rjchapma/etc/evildet.pdf">My evil determinant problem</a>, preprint, 2012.
%H Zhi-Wei Sun, <a href="https://arxiv.org/abs/1308.2900">On some determinants with Legendre symbol entries</a>, arXiv:1308.2900 [math.NT], 2010-2013; Finite Fields Appl. 56(2019), 285-307.
%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 M. Vseminov, <a href="http://dx.doi.org/10.4064/aa159-4-3">On R. Chapman's "evil determinant": case p == 1 (mod 4)</a>, Acta Arith. 159(2013), 331-344.
%e a(2) = 0 since the Legendre symbol ((1-1)/3) is zero.
%t a[n_]:=Det[Table[JacobiSymbol[j-i,Prime[n]],{i,1,(Prime[n]-1)/2},{j,1,(Prime[n]-1)/2}]]
%t Table[a[n],{n,2,20}]
%o (PARI) a(n) = my(p=prime(n)); matdet(matrix((p-1)/2, (p-1)/2, i, j, i--; j--; kronecker(i-j, p))); \\ _Michel Marcus_, Aug 25 2021
%Y Cf. A226163, A227609, A227968, A227971, A228005.
%Y Cf. A094049.
%K sign
%O 2,5
%A _Zhi-Wei Sun_, Aug 09 2013
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