A226163 - OEIS

%I #17 Oct 06 2017 12:08:57

%S 0,-1,0,0,-8,-72,0,0,-2061248,0,-18150912,2581719040,0,0,

%T 6237406973952,0,311692729699401728,0,0,2675112340760315428864,0,0,

%U -149670892669766097645487521792,162894623351898578070944297779200,273248864699809403831952842162176,0,0,-13518055482368485085619549462056665088,4364947372586985974930810143672643878912

%N Determinant of the (p_n-1)/2-by-(p_n-1)/2 matrix with (i,j)-entry being the Legendre symbol ((i^2-((p_n-1)/2)!*j)/p_n), where p_n is the n-th prime.

%C Conjecture: a(n) = 0 if and only if p_n == 3 (mod 4).

%C Note that for an odd prime p we have (((p-1)/2)!)^2 == (-1)^{(p+1)/2} (mod p) by Wilson's theorem. In 1961, Mordell proved that((p-1)/2)! == (-1)^{(h(-p)+1)/2} (mod p) for any prime p > 3 with p == 3 (mod 4), where h(-p) is the class number of the imaginary quadratic field Q(sqrt(-p)).

%H Zhi-Wei Sun, <a href="/A226163/b226163.txt">Table of n, a(n) for n = 2..80</a>

%H L. J. Mordell, <a href="http://www.jstor.org/stable/2312481">The congruence ((p-1)/2)! == 1 or -1 (mod p)</a>, Amer. Math. Monthly 68 (1961), 145-146.

%H Zhi-Wei Sun, <a href="http://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;d4d29df6.1307">A conjecture on Legendre symbol determinants</a>, a message to Number Theory List, July 17, 2013.

%e a(2) = 0 since the Legendre symbol ((1^2-1)/3) is equal to 0.

%t a[n_]:=Det[Table[JacobiSymbol[i^2-((Prime[n]-1)/2)!*j,Prime[n]],{i,1,(Prime[n]-1)/2},{j,1,(Prime[n]-1)/2}]]

%t Table[a[n],{n,2,30}]

%Y Cf. A227609, A227968, A227971.

%K sign

%O 2,5

%A _Zhi-Wei Sun_, Aug 05 2013