A226163 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.

0, -1, 0, 0, -8, -72, 0, 0, -2061248, 0, -18150912, 2581719040, 0, 0, 6237406973952, 0, 311692729699401728, 0, 0, 2675112340760315428864, 0, 0, -149670892669766097645487521792, 162894623351898578070944297779200, 273248864699809403831952842162176, 0, 0, -13518055482368485085619549462056665088, 4364947372586985974930810143672643878912

Offset: 2

Author

Zhi-Wei Sun, Aug 05 2013

Keywords

• sign

Comments

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

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)).

Examples

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

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 2..80
• L. J. Mordell, The congruence ((p-1)/2)! == 1 or -1 (mod p), Amer. Math. Monthly 68 (1961), 145-146.
• Zhi-Wei Sun, A conjecture on Legendre symbol determinants, a message to Number Theory List, July 17, 2013.

Crossrefs

Cf. A227609, A227968, A227971.

Programs

Mathematica

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