A228379 Determinant of the n X n matrix with (i,j)-entry equal to (i^2+j^2)^n for all i,j = 0,...,n-1.

0, -1, -17280, 1168415539200, 980041972344422400000000, -24517645963803990318633839493120000000000, -37442952699741306217982755284947059704721771069440000000000000

Offset: 1

Author

Zhi-Wei Sun, Aug 21 2013

Keywords

• sign

Comments

Conjecture: If n > 2, then (-1)^{n(n-1)/2}*a(n) > 0 and 2*Product_{k=1..n} (k!*(2k-1)!) divides a(n).

This conjecture implies that if p = 2*n-1 > 3 is a prime then we have a((p+1)/2) == 0 (mod p). So Sun's Conjecture 3.7 in the reference would follow from the above conjecture.

If b(n) denotes the n X n determinant with (i,j)-entry equal to (i+j)^n for all i,j = 0,...,n-1, then we conjecture that b(n)*(-1)^(n*(n-1)/2) / ((n-2)!*n*Product_{k=1..n} (k!)) is a positive integer for any integer n > 2.

Links

• Darij Grinberg, Zhi-Wei Sun, and Lilu Zhao, Proof of three conjectures on determinants related to quadratic residues, arXiv:2007.06453 [math.NT], 2020.
• Zhi-Wei Sun, On some determinants with Legendre symbol entries, preprint, arXiv:1308.2900 [math.NT], 2013-2019.

Crossrefs

Cf. A227609.

Programs

Mathematica

a[n_]:=Det[Table[(i^2+j^2)^n,{i,0,n-1},{j,0,n-1}]]
Table[a[n],{n,1,7}]

PARI

a(n) = matdet(matrix(n, n, i, j, ((i-1)^2+(j-1)^2)^n)); \\ Michel Marcus, Jul 13 2020