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A179072
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Chapman's "evil" determinants II.
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6
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-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
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OFFSET
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2,2
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COMMENTS
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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
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REFERENCES
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Richard Guy, Unsolved Problems in Number Theory, 3rd ed., Springer, 2004, Section F5.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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a[n_] := Module[{p, k}, p = Prime[n]; k = (p-1)/2; Det @ Table[JacobiSymbol[ i + j, p], {i, 1, k}, {j, 1, k}]];
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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