A179071 -id:A179071 - cp's OEIS frontend

A179073 A179071 for p == 1 (mod 4).

-2, -18, -4, -70, -882, -32, -182, -29718, -1068, -500, -5604, -4030, -8890182, -776, -1744, -113582, -4832118, -1118, -1111225770, -1764132, -11018, -20000849130, -23156, -71011068, -16432, -2205718, -8920484118, -1063532

Offset: 1

Jonathan Sondow and Wadim Zudilin, Jun 29 2010

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It appears that a(n) is negative and even. (Added Aug 28 2011: This conjecture has been proved by Maxim Vsemirnov.)

M. Vsemirnov, On R. Chapman's "evil determinant": case p == 1 (mod 4), arXiv preprint arXiv:1108.4031 [math.NT], 2011-2012.

Cf. A179071, A179072, A179074.

a71[p_] := Module[{k}, k = (p-1)/2; Det @ Table[JacobiSymbol[j-i, p], {i, 1, k+1}, {j, 1, k+1}]];
a71 /@ Select[Range[1, 301, 4], PrimeQ] (* Jean-François Alcover, Dec 05 2018 *)

A227609 Determinant of the (p_n-1)/2 X (p_n-1)/2 matrix with (i,j)-entry being the Legendre symbol((i^2+j^2)/p_n), where p_n is the n-th prime.

Original entry on oeis.org

-1, 1, -4, -16, -27, 441, -1024, -1024, 34445, -13778944, 82719025, 48841786125, -67649929216, -564926611456, -153908556861703, -25481517249593344, 2456184022341328125, -399780402627654713344, -14448269983744, -214168150727821285287075

Offset: 2

Zhi-Wei Sun, Jul 17 2013

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Conjecture: p_n never divides a(n), and moreover -a(n) is a quadratic residue mod p_n.
Zhi-Wei Sun also made the following conjecture:
  Let p be any odd prime. For each integer d let S(d,p) be the determinant of the (p-1)/2 X (p-1)/2 matrix whose (i,j)-entry is the Legendre symbol ((i^2+d*j^2)/p). If d is a quadratic residue mod p, then so is -S(d,p). If d is a quadratic non-residue mod p, then we have S(d,p) = 0.
These were proved in version 9 of arXiv:1308.2900 (2018). In addition, the author has the following new conjecture.
Conjecture: For any prime p == 3 (mod 4), the number -S(1,p) is a positive square divisible by 2^((p-3)/2), i.e., -S(1,p) = (2^((p-3)/4)*m)^2 for some positive integer m. - Zhi-Wei Sun, Sep 09 2018

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

Alois P. Heinz, Table of n, a(n) for n = 2..100
Zhi-Wei Sun, On some determinants with Legendre symbol entries, preprint, arXiv:1308.2900 [math.NT], 2013-2019.
Zhi-Wei Sun, Is -det[Legendre(i^2+j^2,p)]_{i,j=1,...,(p-1)/2} always a square for each prime p == 3 (mod 4)?, Question 310192 in MathOverflow, Sept. 9, 2018.

Cf. A179071, A179072, A227968, A227971.

with(numtheory): with(LinearAlgebra):
a:= n-> Determinant(Matrix((ithprime(n)-1)/2, (i, j)->
        jacobi(i^2+j^2, ithprime(n)))):
seq(a(n), n=2..20);  # Alois P. Heinz, Jul 18 2013

a[n_]:=Det[Table[JacobiSymbol[i^2+j^2,Prime[n]],{i,1,(Prime[n]-1)/2},{j,1,(Prime[n]-1)/2}]]
Table[a[n],{n,2,20}]

a(n) = my(p=prime(n)); matdet(matrix((p-1)/2, (p-1)/2, i, j, kronecker(i^2+j^2, p))); \\ Michel Marcus, Aug 25 2021

A179072 Chapman's "evil" determinants II.

Original entry on oeis.org

-1, -2, 0, 0, -32, 256, 0, 0, -8192, 0, -262144, 5242880, 0, 0, -33554432, 0, -2684354560, 0, 0, 8589934592000, 0, 0, 932385860354048, 160159261748363264, -1125899906842624, 0, 0, -225179981368524800, 5260204364768739328, 0, 0

Offset: 2

Jonathan Sondow and Wadim Zudilin, Jun 29 2010

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Determinant of the k-by-k matrix with (i,j)-entry L((i+j)/p), where L(./p) denotes the Legendre symbol modulo p and p = p_n = 2k+1 is the n-th prime.
Guy says "Chapman has a number of conjectures which concern the distribution of quadratic residues." One is that if 3 < p_n == 3 (mod 4), then a(n) = 0.
It appears that a(n) is even, if p_n == 1 (mod 4).
For any odd prime p, (p+1)/2-i+(p+1)/2-j == -(i+j-1) (mod p) and hence we have L(-1/p)*|L((i+j)/p)|{i,j=1,...,(p-1)/2} = |L((i+j-1)/p)|{i,j=1,...,(p-1)/2}. Thus the value of a(n) was actually determined in the first reference of R. Chapman. - Zhi-Wei Sun, Aug 21 2013

			p_4 = 7 = 2*3 + 1 and the 3 X 3 matrix (L((i+j)/7)) is
   1, -1,  1
  -1,  1, -1
   1, -1, -1
which has determinant 0, so a(4) = 0.

Richard Guy, Unsolved Problems in Number Theory, 3rd ed., Springer, 2004, Section F5.

Robin Chapman, Determinants of Legendre symbol matrices, Acta Arith. 115 (2004), 231-244.
Robin Chapman, Steinitz classes of unimodular lattices, European J. Combin. 25 (2004), 487-493.
Robin Chapman, My evil determinant problem, 2009.
Maxim Vsemirnov, On R. Chapman's "evil determinant": case p=1 (mod 4), arXiv:1108.4031 [math.NT], 2011-2012.
Maxim Vsemirnov, On the evaluation of R. Chapman's "evil determinant", Linear Algebra Appl. 436(2012), 4101-4106.
Wikipedia, Legendre symbol

Cf. A179071 (Chapman's "evil" determinants I), A179073 (A179071 for p == 1 (mod 4)), A179074 (A179072 for p == 1 (mod 4)).

a[n_] := Module[{p, k}, p = Prime[n]; k = (p-1)/2; Det @ Table[JacobiSymbol[ i + j, p], {i, 1, k}, {j, 1, k}]];
Table[a[n], {n, 2, 32}] (* Jean-François Alcover, Nov 18 2018 *)

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.

Original entry on oeis.org

-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

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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

A179074 A179072 for p == 1 (mod 4).

Original entry on oeis.org

-2, -32, 256, -8192, -262144, 5242880, -33554432, -2684354560, 8589934592000, 932385860354048, 160159261748363264, -1125899906842624, -225179981368524800, 5260204364768739328, 43977037871723571052544

Offset: 1

Jonathan Sondow and Wadim Zudilin, Jun 29 2010

sign

It appears that a(n) is even.

Cf. A179071, A179072, A179073.

a72[p_] := Module[{k}, k = (p-1)/2; Det @ Table[JacobiSymbol[i+j, p], {i, 1, k}, {j, 1, k}]];
a72 /@ Select[Range[1, 201, 4], PrimeQ] (* Jean-François Alcover, Dec 05 2018 *)

A372385 Determinant of the matrix [Jacobi(i - j, 2n + 1)]_{0 < i, j < n}, where Jacobi(a, m) denotes the Jacobi symbol (a/m).

Original entry on oeis.org

0, 1, 2, 1, 8, 16, 2, 1, -56, 1, 0, 0, 44, 1, 127776, 1067089, 592, 7311616, 22, 1, 0, 1, 0, 98867482624, 132, 1933242748921, 31578700795392, 1, 22108, 0, -6001552839958, 1, -2401383183590221824, 1, 818, 0, 95506686981729056, 1, 0, 1, 30328979503109918400, 208688450689382571638784, 394

Offset: 2

Zhi-Wei Sun, Apr 29 2024

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In a 2003 preprint, R. Chapman conjectured that if 2*n + 1 (with n > 1) is a prime congruent to 3 modulo 4, then a(n) = 1.
Conjecture 1: If p = 2*n + 1 is a prime congruent to 1 modulo 4, then det[x + Jacobi(i - j, p)]{0 < i, j < n} = r_p - 2*s_p + (-1)^{n/2}*x*(2*r_p - p*s_p), and det[x + Jacobi(i - j, p)]{0 <= i, j <= n} = -r_p + (-1)^{n/2}*p*s_p*x, where r_p and s_p are rational numbers such that e_p^{(2 - Jacobi(2, p))*h(p)} = r_p + s_p*sqrt(p), e_p and h(p) are the fundamental unit and the class number of the real quadratic field Q(sqrt(p)), respectively.
Conjecture 2: Suppose that p = 2*n + 1 is a prime greater than 3, and write e_p^{h(p)} = a_p + b_p*sqrt(p) with 2*a_p and 2*b_p integral.
(1) det[x + Jacobi(i + j, p)]_{0 < i, j < n} is (-1)^{n/2}*2^{n - 1}*(b_p - a_p*x) if p == 1 (mod 4), and 2^{n-1}*x if p == 3 (mod 4).
(2) det[x + Jacobi(i + j, p)]_{0 <= i, j < n} is (-1)^{n/2}*2^{n-1}*(2*b_p - a_p + x*(p*b_p - 2*a_p)) if p == 1 (mod 4), and -2^{n - 1}*(2*x + 1) if p == 3 (mod 4).
We have checked both conjectures for n up to 1000.

			a(3) = 1 since the determinant of the matrix [Jacobi(i-j, 2*3+1)]_{0 < i, j < 3} = [0, -1; 1, 0] has the value 1.

Zhi-Wei Sun, Table of n, a(n) for n = 2..144
R. Chapman, Determinants of Legendre symbol matrices, preprint, 2003. (Cf. Conjecture 2 of this paper.)
R. Chapman, Determinants of Legendre symbol matrices, Acta Arith. 115 (2004), 231-244.
Z.-W. Sun, Conjectural values of some determinants involving Legendre symbols (I), Question 470324 at MathOverflow, April 30, 2024.
M. Vsemirnov, On the evaluation of R. Chapman's "evil determinant", Linear Algebra Appl. 436 (2012), 4101-4106.
M. Vsemirnov, On R. Chapman's "evil determinant": case p == 1(mod 4), Acta Arith. 159 (2013), 331-344; see also the arXiv version, arXiv:1108.4031 [math.NT], 2011-2012.

Cf. A179071, A179072, A372314.

a[n_]:=a[n]=Det[Table[JacobiSymbol[i-j, 2n+1], {i, 1, n-1}, {j, 1, n-1}]];
tab={}; Do[tab=Append[tab, a[n]], {n, 2, 44}]; Print[tab]

a(n) = matdet(matrix(n-1, n-1, i, j, kronecker(i-j, 2*n+1))); \\ Michel Marcus, Apr 29 2024

from sympy import Matrix, jacobi_symbol
def A372385(n): return Matrix(n-1,n-1,[jacobi_symbol(i-j,(n<<1)|1) for i in range(n-1) for j in range(n-1)]).det() # Chai Wah Wu, Apr 29 2024

A372409 Determinant of the matrix obtained from [Jacobi(i-j, 2*n+1)]_{0

Original entry on oeis.org

1, -1, 1, -3, 3, -8, -5, -5, -24, 7, 0, 0, 9, 9, -81312, -1341867, 11, -19685120, -13, -13, 0, -15, 0, -180287762432, 17, -1407939911477, 10526233598464, 19, 19, 0, 6040299856799, -21, 29830847001120768, 23, -23, 0, 115407361849089836, -25, 0, 27, 104060523591574200

Offset: 2

Zhi-Wei Sun, Apr 29 2024

sign

Conjecture 1: If 2*n + 1 (with n > 1) is a prime p, then a(n)/floor((p-2)/3) coincides with (-1)^((p+3)/4) if p == 1 (mod 4), and (-1)^((h(-p)+1)/2) if p == 3 (mod 4), where h(-p) is the class number of the imaginary quadratic field Q(sqrt(-p)).
Conjecture 2: Let p > 3 be a prime, and let S(p) and T(p) denote the matrices obtained from [Jacobi(i+j,p)]{1<=i,j<=(p-3)/2} and [Jacobi(i+j,p)]{0<=i,j<=(p-3)/2} (respectively) by replacing all the entries in the first row by 1. Then det S(p) = -det T(p) = 2^((p-5)/2)*s(p), where s(p) is (-1)^((p+3)/4) if p == 1 (mod 4), and (-1)^((h(-p)-1)/2) if p == 3 (mod 4).
Both conjectures are motivated by Conjecture 4.6 in the author's 2019 FFA paper as well as the conjectures in A372385. They have been verified for primes p < 2000.

			a(3) = -1 since the determinant of the matrix [1, 1; Jacobi(2-1,2*3+1), Jacobi(2-2,2*3+1)] = [1, 1; 1, 0] has the value -1.

L.-Y. Wang and H.-L. Wu, On certain determinants involving Legendre symbols, Ramanujan J. 58 (2022), 43-56.

R. Chapman, Determinants of Legendre symbol matrices, Acta Arith. 115 (2004), 231-244.
Z.-W. Sun, On some determinants with Legendre symbol entries, Finite Fields Appl. 56 (2019), 285-307.
M. Vsemirnov, On the evaluation of R. Chapman's "evil determinant", Linear Algebra Appl. 436 (2012), 4101-4106.
M. Vsemirnov, On R. Chapman's "evil determinant": case p == 1(mod 4), Acta Arith. 159 (2013), 331-344; see also the arXiv version, arXiv:1108.4031 [math.NT], 2011-2012.
L.-Y. Wang and H.-L. Wu, On p-th cyclotomic field and Zhi-Wei Sun's Legendre determinants over F_p, arXiv:2401.05853 [math.NT], 2024.

Cf. A179071, A179072, A372385.

a[n_]:=a[n]=Det[Table[If[i==1,1,JacobiSymbol[i-j,2*n+1]],{i,1,n-1},{j,1,n-1}]];
tab={};Do[tab=Append[tab,a[n]],{n,2,42}];Print[tab]

a(n) = matdet(matrix(n-1, n-1, i, j, if (i==1, 1, kronecker(i-j, 2*n+1)))); \\ Michel Marcus, Apr 30 2024

from sympy import Matrix, jacobi_symbol
def A372409(n): return Matrix(n-1,n-1,[jacobi_symbol(i-j,(n<<1)|1) if i else 1 for i in range(n-1) for j in range(n-1)]).det() # Chai Wah Wu, May 01 2024

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A179073 A179071 for p == 1 (mod 4).

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A227609 Determinant of the (p_n-1)/2 X (p_n-1)/2 matrix with (i,j)-entry being the Legendre symbol((i^2+j^2)/p_n), where p_n is the n-th prime.

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A179072 Chapman's "evil" determinants II.

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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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A179074 A179072 for p == 1 (mod 4).

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A372385 Determinant of the matrix [Jacobi(i - j, 2n + 1)]_{0 < i, j < n}, where Jacobi(a, m) denotes the Jacobi symbol (a/m).

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A372409 Determinant of the matrix obtained from [Jacobi(i-j, 2*n+1)]_{0

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