A179073 -id:A179073 - cp's OEIS frontend

A094048 Let p(n) be the n-th prime congruent to 1 mod 4. Then a(n) = the least m for which m^2+1=p(n)*k^2 has a solution.

2, 18, 4, 70, 6, 32, 182, 29718, 1068, 500, 5604, 10, 8890182, 776, 1744, 113582, 4832118, 1118, 1111225770, 1764132, 14, 1710, 23156, 71011068, 16, 82, 8920484118, 1063532, 2482, 126862368, 352618

Offset: 1

Matthijs Coster, Apr 29 2004

nonn

Subsequence of A191860. [Reinhard Zumkeller, Jun 18 2011]

Cf. A002144, A094049 (associated k), A130226, A137351, A179073.

Cf. A000196, A037213, A000290.

a094048 n = head [m | m <- map (a037213 . subtract 1 . (* a002144 n))
                               (tail a000290_list), m > 0]
-- Reinhard Zumkeller, Jun 13 2015

f[n_] := Block[{y = 1}, While[ !IntegerQ[ Sqrt[n*y^2 - 1]], y++ ]; Sqrt[n*y^2 - 1]]; lst = {}; Do[p = Prime@ n; If[ Mod[p, 4] == 1, AppendTo[lst, f@p]; Print[{n, Prime@n, f@p}]], {n, 66}]; lst

Edited by Don Reble, Apr 30 2004

A179071 Chapman's "evil" determinants I.

Original entry on oeis.org

1, -2, 1, 1, -18, -4, 1, 1, -70, 1, -882, -32, 1, 1, -182, 1, -29718, 1, 1, -1068, 1, 1, -500, -5604, -4030, 1, 1, -8890182, -776, 1, 1, -1744, 1, -113582, 1, -4832118, 1, 1, -1118, 1, -1111225770, 1, -1764132, -11018, 1, 1, 1, 1, -20000849130, -23156, 1

Offset: 2

Jonathan Sondow and Wadim Zudilin, Jun 29 2010

sign

Determinant of the (k+1)X(k+1) matrix with (i,j)-entry L((j-i)/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 p_n == 3 (mod 4), then a(n) = 1. Chapman also has a conjecture if p_n == 1 (mod 4), involving the fundamental unit and class number of the quadratic field Q(sqrt(p)). (Added Aug 23 2011: Both conjectures have been proved by Vsemirnov.)
It appears that a(n) is negative and even, if p_n == 1 (mod 4); see A179073. (Added Aug 28 2011: This conjecture has also been proved by Vsemirnov.)

			p_3 = 5 = 2*2+1 and the (2+1)-by-(2+1) matrix (L((j-i)/5)) is
   0,  1, -1
   1,  0,  1
  -1,  1,  0
which has determinant -2, so a(3) = -2.

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. A179072 (Chapman's "evil" determinants II), 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[ j-i, p], {i, 1, k+1}, {j, 1, k+1}]];
Table[a[n], {n, 2, 52}] (* Jean-François Alcover, Nov 18 2018 *)

a(n) = my(p=prime(n), k=(p+1)/2); matdet(matrix(k, k, i, j, kronecker(j-i, 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

sign

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 *)

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 *)

cp's OEIS Frontend

A094048 Let p(n) be the n-th prime congruent to 1 mod 4. Then a(n) = the least m for which m^2+1=p(n)*k^2 has a solution.

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A179071 Chapman's "evil" determinants I.

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A179072 Chapman's "evil" determinants II.

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

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