A176113 -id:A176113 - cp's OEIS frontend

A227971 Determinant of the (p_n+1)/2 X (p_n+1)/2 matrix with (i,j)-entry (i,j=0,...,(p_n-1)/2) being the Legendre symbol((i+j)/p_n), where p_n is the n-th prime.

-1, 2, 8, 32, 96, -1024, 512, 2048, 40960, 32768, 1572864, -33554432, 2097152, 8388608, 234881024, 536870912, 20937965568, 8589934592, 34359738368, -73392401154048, 549755813888, 2199023255552, -8796093022208000, -1577385769486516224, 11258999068426240

Offset: 2

Zhi-Wei Sun, Aug 01 2013

sign

Conjecture: If p_n == 1 (mod 4), then a(n) == ((p_n-1)/2)! (mod p_n). If p_n == 3 (mod 4), then a(n) == (2/p_n) (mod p_n).
Zhi-Wei Sun also made the following general conjecture:
Let p be any odd prime. For each integer d let R(d,p) be the determinant of the (p+1)/2-by-(p+1)/2 matrix whose (i,j)-entry (i,j = 0,...,(p-1)/2) is the Legendre symbol ((i+d*j)/p). When p == 3 (mod 4), we have R(d,p) == (2/p) (mod p) if (d/p) = 1, and R(d,p) == 1 (mod p) if (d/p) = -1. In the case p == 1 (mod 4), we have R(c^2*d,p) == (c/p)*R(d,p) (mod p) for any integer c, and R(d,p) == 1 or -1 (mod p) if (d/p) = -1.
The author could prove that for any odd prime p and integer d not divisible by p, the determinant of the (p-1)-by-(p-1) matrix with (i,j)-entry (i,j=1,...,p-1) being the Legendre symbol ((i+dj)/p) has the exact value (-d/p)*p^{(p-3)/2}.
On August 19 2013, Zhi-Wei Sun found a formula for a(n). Namely, he made the following conjecture: If p_n == 1 (mod 4) and e(p_n)^{h(p_n)} = (a_n + b_n*sqrt(p_n))/2 with a_n and b_n integers of the same parity (where e(p_n) and h(p_n) are the fundamental unit and the class number of the quadratic field Q(sqrt(p_n)) respectively), then a(n) = - (2/p_n)*2^{(p_n-3)/2}*a_n. If p_n > 3 and p_n == 3 (mod 4), then  a(n) = 2^{(p_n-1)/2}.
On August 19 2013, Zhi-Wei Sun proved all the conjectured congruences mentioned above by using the identity D(c,d,n) = (-d)^{n*(n+1)/2}*(n!)^{n+1}, where D(c,d,n) is the (n+1) X (n+1) determinant with (i,j)-entry equal to (i+d*j+c)^n for all i,j = 0,...,n. For any prime p == 1 (mod 4) he showed that R(d,p) == (d*(d/p))^{(p-1)/4}*((p-1)/2)! (mod p). Note also that the formula for a(n) found by Sun on August 9, 2013 is actually equivalent to Chapman's result on the evaluation of the determinant |((i+j-1)/p)|_{i,j=1,...,(p+1)/2}.

			a(2) = -1 since the determinant |((i+j)/3)|_{i=0,1; j=0,1}| equals -1.

Zhi-Wei Sun, Table of n, a(n) for n = 2..100
L. Carlitz, Some cyclotomic matrices, Acta Arith. 5(1959), 293-308.
Robin Chapman, Determinants of Legendre symbol matrices, Acta Arith. 115 (2004), 231-244.
Zhi-Wei Sun, A conjecture on Legendre symbol determinants, a message to Number Theory List, July 17, 2013.
Zhi-Wei Sun, On some determinants with Legendre symbol entries, arXiv:1308.2900 [math.NT], 2010-2013; Finite Fields Appl. 56(2019), 285-307,
Maxim Vsemirnov, On the evaluation of R. Chapman's "evil determinant", Linear Algebra Appl. 436(2012), 4101-4106.
Maxim Vsemirnov, On R. Chapman's "evil determinant": case p == 1 (mod 4), Acta Arith. 159 (2013), 331-344.

Cf. A176113, A179071, A179072, A226163, A227609, A227968, A228005, A228077.

a[n_] := Det[Table[JacobiSymbol[i+j, Prime[n]], {i, 0, (Prime[n]-1)/2}, {j, 0, (Prime[n]-1)/2}]]; Table[a[n], {n, 2, 30}]

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

A228252 Determinant of the (n+1) X (n+1) matrix with (i,j)-entry equal to (i-2j)^n for all i,j = 0,...,n.

Original entry on oeis.org

1, 2, 64, 82944, 8153726976, 97844723712000000, 210357201231685877760000000, 111759427954264225978066246041600000000, 19353724511515955943723861007628909886308352000000000, 1393093075882582456065167957036969287436705021776979747143680000000000, 51765823014530203817669442380756522498563227474168874049894256476160000000000000000000000

Offset: 0

Zhi-Wei Sun, Aug 19 2013

nonn

Note that a(n) = D(n,n,-2,0), where D(k,n,x,y) denotes the (n+1) X (n+1) determinant with (i,j)-entry equal to (i+j*x+y)^k for all i,j = 0,...,n. By the comments in A176113, it is known that D(n,n,x,y) = (-x)^{n*(n+1)/2}*(n!)^{n+1}. Note also that D(k,n,x,y) = 0 for all k = 0,...,n-1, which can be proved by using the definition of determinant and the binomial theorem.

For any matrices M of this pattern, M(i, j) = M(i-2, j-1). - Iain Fox, Feb 26 2018

			Northwest corner of matrix corresponding to a(n):
0^n  (-2)^n  (-4)^n  (-6)^n  (-8)^n
  1  (-1)^n  (-3)^n  (-5)^n  (-7)^n
2^n       0  (-2)^n  (-4)^n  (-6)^n
3^n       1  (-1)^n  (-3)^n  (-5)^n
4^n     2^n       0  (-2)^n  (-4)^n

J. M. Monier, Algèbre et géometrie, Dunod, 1996.

Iain Fox, Table of n, a(n) for n = 0..28
C. Krattenthaler, Advanced Determinant Calculus: A Complement, Linear Algebra Appl. 411 (2005), 68-166; arXiv:math/0503507 [math.CO], 2017.

Cf. A176113.

a[n_]:=Det[Table[If[n==0,1,(i-2j)^n],{i,0,n},{j,0,n}]]
Table[a[n],{n,0,10}]

a(n) = matdet(matrix(n+1, n+1, i, j, (i - 2*j + 1)^n)) \\ Iain Fox, Feb 16 2018

a(n) = 2^(n*(n+1)/2)*(n!)^(n+1) \\ (faster and uses less memory) Iain Fox, Apr 15 2018

a(n) = 2^(n*(n+1)/2)*(n!)^(n+1) as shown by comments. - Iain Fox, Apr 15 2018

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A227971 Determinant of the (p_n+1)/2 X (p_n+1)/2 matrix with (i,j)-entry (i,j=0,...,(p_n-1)/2) being the Legendre symbol((i+j)/p_n), where p_n is the n-th prime.

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A228252 Determinant of the (n+1) X (n+1) matrix with (i,j)-entry equal to (i-2j)^n for all i,j = 0,...,n.

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